Calculus/Differentiation/Basics of Differentiation/Solutions

Find the Derivative by Definition

1.

f(x)=x^{2}\,

{\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{2}-x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {x^{2}+2x\Delta x+\Delta x^{2}-x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x\Delta x+\Delta x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}2x+\Delta x\\&=\mathbf {2x} \end{aligned}}

2.

f(x)=2x+2\,

{\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {[2(x+\Delta x)+2]-(2x+2)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x+2\Delta x+2-2x-2}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x}}\\&=\mathbf {2} \end{aligned}}

3.

f(x)={\frac {1}{2}}x^{2}\,

{\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{2}}(x+\Delta x)^{2}-{\frac {1}{2}}x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{2}}(x^{2}+2x\Delta x+\Delta x^{2})-{\frac {1}{2}}x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {{\frac {x^{2}}{2}}+{\frac {2x\Delta x}{2}}+{\frac {\Delta x^{2}}{2}}-{\frac {x^{2}}{2}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x\Delta x+\Delta x^{2}}{2\Delta x}}\\&=\lim _{\Delta x\to 0}x+{\frac {\Delta x}{2}}\\&=\mathbf {x} \end{aligned}}

4.

f(x)=2x^{2}+4x+4\,

{\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {[2(x+\Delta x)^{2}+4(x+\Delta x)+4]-(2x^{2}+4x+4)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {4x\Delta x+2\Delta x^{2}+4\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0}4x+2\Delta x+4\\&=\mathbf {4x+4} \end{aligned}}

5.

f(x)={\sqrt {x+2}}\,

{\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\sqrt {x+\Delta x+2}}-{\sqrt {x+2}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}({\frac {{\sqrt {x+\Delta x+2}}-{\sqrt {x+2}}}{\Delta x}})({\frac {{\sqrt {x+\Delta x+2}}+{\sqrt {x+2}}}{{\sqrt {x+\Delta x+2}}+{\sqrt {x+2}}}})\\&=\lim _{\Delta x\to 0}{\frac {x+\Delta x+2-x-2}{\Delta x({\sqrt {x+\Delta x+2}}+{\sqrt {x+2}})}}\\&=\lim _{\Delta x\to 0}{\frac {\Delta x}{\Delta x({\sqrt {x+\Delta x+2}}+{\sqrt {x+2}})}}\\&=\lim _{\Delta x\to 0}{\frac {1}{{\sqrt {x+\Delta x+2}}+{\sqrt {x+2}}}}\\&=\mathbf {\frac {1}{2{\sqrt {x+2}}}} \end{aligned}}

6.

f(x)={\frac {1}{x}}\,

{\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{x+\Delta x}}-{\frac {1}{x}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\frac {x-x-\Delta x}{x(x+\Delta x)}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {-\Delta x}{x\Delta x(x+\Delta x)}}\\&=\lim _{\Delta x\to 0}{\frac {-1}{x(x+\Delta x)}}\\&=\mathbf {-{\frac {1}{x^{2}}}} \end{aligned}}

7.

f(x)={\frac {3}{x+1}}\,

{\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {3}{x+\Delta x+1}}-{\frac {3}{x+1}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\frac {3x+3-(3x+3\Delta x+3)}{(x+1)(x+\Delta x+1)}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {-3\Delta x}{\Delta x(x+1)(x+\Delta x+1)}}\\&=\lim _{\Delta x\to 0}{\frac {-3}{(x+1)(x+\Delta x+1)}}\\&=\mathbf {\frac {-3}{(x+1)^{2}}} \end{aligned}}

8.

f(x)={\frac {1}{\sqrt {x+1}}}\,

{\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{\sqrt {x+\Delta x+1}}}-{\frac {1}{\sqrt {x+1}}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\frac {{\sqrt {x+1}}-{\sqrt {x+\Delta x+1}}}{{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}({\frac {{\sqrt {x+1}}-{\sqrt {x+\Delta x+1}}}{\Delta x{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}}})({\frac {{\sqrt {x+1}}+{\sqrt {x+\Delta x+1}}}{{\sqrt {x+1}}+{\sqrt {x+\Delta x+1}}}})\\&=\lim _{\Delta x\to 0}{\frac {x+1-(x+\Delta x+1)}{\Delta x{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}({\sqrt {x+\Delta x+1}}+{\sqrt {x+1}})}}\\&=\lim _{\Delta x\to 0}{\frac {-1}{{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}({\sqrt {x+\Delta x+1}}+{\sqrt {x+1}})}}\\&={\frac {-1}{(x+1)(2{\sqrt {x+1}})}}\\&=\mathbf {\frac {-1}{2(x+1)^{3/2}}} \end{aligned}}

9.

f(x)={\frac {x}{x+2}}\,

{\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {x+\Delta x}{x+\Delta x+2}}-{\frac {x}{x+2}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)(x+2)-x(x+\Delta x+2)}{\Delta x(x+\Delta x+2)(x+2)}}\\&=\lim _{\Delta x\to 0}{\frac {x^{2}+2x+x\Delta x+2\Delta x-x^{2}-x\Delta x-2x}{\Delta x(x+\Delta x+2)(x+2)}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x(x+\Delta x+2)(x+2)}}\\&=\lim _{\Delta x\to 0}{\frac {2}{(x+\Delta x+2)(x+2)}}\\&=\mathbf {\frac {2}{(x+2)^{2}}} \end{aligned}}

Prove the Constant Rule

10. Use the definition of the derivative to prove that for any fixed real number

c

,

{\frac {d}{dx}}\left[cf(x)\right]=c{\frac {d}{dx}}\left[f(x)\right]

{\begin{aligned}{\frac {d}{dx}}\left[cf(x)\right]&=\lim _{\Delta x\to 0}{\frac {cf\left(x+\Delta x\right)-cf\left(x\right)}{\Delta x}}\\&=c\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\\&=c{\frac {d}{dx}}\left[f(x)\right]\end{aligned}}

Find the Derivative by Rules

Power Rule

11.

f(x)=2x^{2}+4\,

f'(x)=\mathbf {4x}

12.

f(x)=3{\sqrt[{3}]{x}}\,

f'(x)=3({\frac {1}{3}})x^{-2/3}=\mathbf {\frac {1}{\sqrt[{3}]{x^{2}}}}

13.

f(x)=2x^{5}+8x^{2}+x-78\,

f'(x)=\mathbf {10x^{4}+16x+1}

14.

f(x)=7x^{7}+8x^{5}+x^{3}+x^{2}-x\,

f'(x)=\mathbf {49x^{6}+40x^{4}+3x^{2}+2x-1}

15.

f(x)={\frac {1}{x^{2}}}+3x^{\frac {1}{3}}\,

f'(x)={\frac {-2}{x^{3}}}+x^{-2/3}=\mathbf {{\frac {-2}{x^{3}}}+{\frac {1}{\sqrt[{3}]{x^{2}}}}}

16.

f(x)=3x^{15}+{\frac {1}{17}}x^{2}+{\frac {2}{\sqrt {x}}}\,

f'(x)=45x^{14}+{\frac {2}{17}}x-{\frac {1}{\sqrt {x^{3}}}}=\mathbf {45x^{14}+{\frac {2}{17}}x-{\frac {1}{x{\sqrt {x}}}}}

17.

f(x)={\frac {3}{x^{4}}}-{\sqrt[{4}]{x}}+x\,

f'(x)={\frac {-12}{x^{5}}}-{\frac {1}{4}}x^{-3/4}+1=\mathbf {{\frac {-12}{x^{5}}}-{\frac {1}{4{\sqrt[{4}]{x^{3}}}}}+1}

18.

f(x)=6x^{1/3}-x^{0.4}+{\frac {9}{x^{2}}}\,

f'(x)=2x^{-2/3}-0.4x^{-0.6}-{\frac {18}{x^{3}}}=\mathbf {{\frac {2}{\sqrt[{3}]{x^{2}}}}-{\frac {0.4}{x^{0.6}}}-{\frac {18}{x^{3}}}}

19.

f(x)={\frac {1}{\sqrt[{3}]{x}}}+{\sqrt {x}}\,

f'(x)=-{\frac {1}{3x^{4/3}}}+{\frac {1}{2{\sqrt {x}}}}=\mathbf {{\frac {-1}{3x{\sqrt[{3}]{x}}}}+{\frac {1}{2{\sqrt {x}}}}}

Product Rule

20.

f(x)=(x^{4}+4x+2)(2x+3)\,

f'(x)=(4x^{3}+4)(2x+3)+(x^{4}+4x+2)(2)=\mathbf {10x^{4}+12x^{3}+16x+16}

21.

f(x)=(2x-1)(3x^{2}+2)\,

f'(x)=(2)(3x^{2}+2)+(2x-1)(6x)=\mathbf {18x^{2}-6x+4}

22.

f(x)=(x^{3}-12x)(3x^{2}+2x)\,

f'(x)=(3x^{2}-12)(3x^{2}+2x)+(x^{3}-12x)(6x+2)=\mathbf {15x^{4}+8x^{3}-108x^{2}-48x}

23.

f(x)=(2x^{5}-x)(3x+1)\,

f'(x)=(10x^{4}-1)(3x+1)+(2x^{5}-x)(3)=\mathbf {36x^{5}+10x^{4}-6x-1}

Quotient Rule

29.

f(x)={\frac {2x+1}{x+5}}\,

f'(x)={\frac {(x+5)(2)-(2x+1)}{(x+5)^{2}}}=\mathbf {\frac {9}{(x+5)^{2}}}

30.

f(x)={\frac {3x^{4}+2x+2}{3x^{2}+1}}\,

f'(x)={\frac {(3x^{2}+1)(12x^{3}+2)-(3x^{4}+2x+2)(6x)}{(3x^{2}+1)^{2}}}=\mathbf {\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}}

31.

f(x)={\frac {x^{\frac {3}{2}}+1}{x+2}}\,

f'(x)={\frac {(x+2)({\frac {3}{2}}{\sqrt {x}})-(x^{\frac {3}{2}}+1)}{(x+2)^{2}}}=\mathbf {\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}}

32.

d(u)={\frac {u^{3}+2}{u^{3}}}\,

d'(u)={\frac {u^{3}(3u^{2})-(u^{3}+2)(3u^{2})}{u^{6}}}=\mathbf {-{\frac {6}{u^{4}}}}

33.

f(x)={\frac {x^{2}+x}{2x-1}}\,

f'(x)={\frac {(2x-1)(2x+1)-(x^{2}+x)(2)}{(2x-1)^{2}}}=\mathbf {\frac {2x^{2}-2x-1}{(2x-1)^{2}}}

34.

f(x)={\frac {x+1}{2x^{2}+2x+3}}\,

f'(x)={\frac {(2x^{2}+2x+3)-(x+1)(4x+2)}{(2x^{2}+2x+3)^{2}}}=\mathbf {\frac {-2x^{2}-4x+1}{(2x^{2}+2x+3)^{2}}}

35.

f(x)={\frac {16x^{4}+2x^{2}}{x}}\,

f'(x)={\frac {x(64x^{3}+4x)-(16x^{4}+2x^{2})}{x^{2}}}=\mathbf {48x^{2}+2}

Chain Rule

43.

f(x)=(x+5)^{2}\,

Let

f(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x+5

. Then

f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x+5)(1)=\mathbf {2(x+5)}

44.

g(x)=(x^{3}-2x+5)^{2}\,

Let

g(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=x^{3}-2x+5

. Then

g'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=\mathbf {2(x^{3}-2x+5)(3x^{2}-2)}

45.

f(x)={\sqrt {1-x^{2}}}\,

Let

f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1-x^{2}

. Then

f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1-x^{2}}}}}(-2x)=\mathbf {-{\frac {x}{\sqrt {1-x^{2}}}}}

46.

f(x)={\frac {(2x+4)^{3}}{4x^{3}+1}}\,

Let

f(x)={\frac {N(x)}{D(x)}};\quad N(x)=g(h(x));\quad g(x)=x^{3};\quad h(x)=2x+4;\quad D(x)=4x^{3}+1

. Then

$f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}$
$N'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=3(2x+4)^{2}(2)=6(2x+4)^{2}$
$D'(x)=12x^{2}$

f'(x)={\frac {(4x^{3}+1)6(2x+4)^{2}-(2x+4)^{3}12x^{2}}{(4x^{3}+1)^{2}}}=\mathbf {\frac {6(4x^{3}+1)(2x+4)^{2}-12x^{2}(2x+4)^{3}}{(4x^{3}+1)^{2}}}

47.

f(x)=(2x+1){\sqrt {2x+2}}\,

Let

f(x)=A(x)B(x);\quad A(x)=2x+1;\quad B(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x+2

. Then

$f'(x)=A'(x)B(x)+A(x)B'(x)$
$A'(x)=2$
$B'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}$

f'(x)=\mathbf {2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}}

48.

f(x)={\frac {2x+1}{\sqrt {2x+2}}}\,

Let

f(x)={\frac {N(x)}{D(x)}};\quad N(x)=2x+1;\quad D(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x+2

. Then

$f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}$
$N'(x)=2$
$D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}$

f'(x)={\frac {{\sqrt {2x+2}}(2)-{\frac {(2x+1)}{\sqrt {2x+2}}}}{2x+2}}=\mathbf {\frac {2x+3}{(2x+2)^{3/2}}}

49.

f(x)={\sqrt {2x^{2}+1}}(3x^{4}+2x)^{2}\,

Let

f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x^{2}+1;\quad B(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=3x^{4}+2x

. Then

$f'(x)=A'(x)B(x)+A(x)B'(x)$
$A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x^{2}+1}}}}(4x)={\frac {2x}{\sqrt {2x^{2}+1}}}$
$B'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=2(3x^{4}+2x)(12x^{3}+2)$

f'(x)={\frac {2x}{\sqrt {2x^{2}+1}}}(3x^{4}+2x)^{2}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)=\mathbf {{\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)}

50.

f(x)={\frac {2x+3}{(x^{4}+4x+2)^{2}}}\,

Let

f(x)={\frac {N(x)}{D(x)}};\quad N(x)=2x+3;\quad D(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x^{4}+4x+2

. Then

$f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}$
$N'(x)=2$
$D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x^{4}+4x+2)(4x^{3}+4)$

f'(x)={\frac {(x^{4}+4x+2)^{2}(2)-(2x+3)(2)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}=\mathbf {\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}

51.

f(x)={\sqrt {x^{3}+1}}(x^{2}-1)\,

Let

f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=x^{3}+1;\quad B(x)=x^{2}-1

. Then

$f'(x)=A'(x)B(x)+A(x)B'(x)$
$A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {x^{3}+1}}}}(3x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}$
$B'(x)=2x$

f'(x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}(x^{2}-1)+{\sqrt {x^{3}+1}}(2x)=\mathbf {{\frac {3x(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}}

52.

f(x)=((2x+3)^{4}+4(2x+3)+2)^{2}\,

Let

f(x)=g((h(x));\quad g(x)=x^{2};\quad h(x)=A(x)+B(x)+2;\quad A(x)=r(s(x));\quad r(x)=x^{4};\quad s(x)=2x+3;\quad B(x)=4(2x+3)

. Then

$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(A(x)+B(x)+2)(A'(x)+B'(x))$
$A'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=4(2x+3)^{3}(2)=8(2x+3)^{3}$
$B'(x)=8$

f'(x)=\mathbf {2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)}

53.

f(x)={\sqrt {1+x^{2}}}\,

Let

f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1+x^{2}

. Then

f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}(2x)=\mathbf {\frac {x}{\sqrt {1+x^{2}}}}

Exponentials

54.

f(x)=(3x^{2}+e)e^{2x}\,

f'(x)=f'(x)=(6x)e^{2x}+(3x^{2}+e)2e^{2x}=\mathbf {6xe^{2x}+2e^{2x}(3x^{2}+e)}

55.

f(x)=e^{2x^{2}+3x}

Let

f(x)=g(h(x));\quad g(x)=e^{x};\quad h(x)=2x^{2}+3x

. Then

f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=e^{2x^{2}+3x}(4x+3)=\mathbf {(4x+3)e^{2x^{2}+3x}}

56.

f(x)=e^{e^{2x^{2}+1}}

Let

u(x)=e^{x}

v(x)=e^{x}

w(x)=2x^{2}+1

Then

f(x)=u(v(w(x)))

Using the chain rule, we have

{\frac {df}{dx}}={\frac {du}{dv}}{\frac {dv}{dw}}{\frac {dw}{dx}}

The individual factor are

{\frac {du}{dv}}={\frac {d(e^{v})}{dv}}=e^{v}=e^{e^{w}}=e^{e^{2x^{2}+1}}

{\frac {dv}{dw}}={\frac {d(e^{w})}{dw}}=e^{w}=e^{2x^{2}+1}

{\frac {dw}{dx}}={\frac {d(2x^{2}+1)}{dx}}=4x

So

{\frac {df}{dx}}=(e^{e^{2x^{2}+1}})(e^{2x^{2}+1})(4x)=\mathbf {4xe^{2x^{2}+1+e^{2x^{2}+1}}}

57.

f(x)=4^{x}\,

f'(x)=\mathbf {\ln(4)4^{x}}

Logarithms

58.

f(x)=2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+\ln x\,

f'(x)=\ln(2)2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+2^{x-3}{\frac {3}{2{\sqrt {x^{3}-2}}}}3x^{2}+{\frac {1}{x}}=\mathbf {3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-4}}{\sqrt {x^{3}-2}}}+{\frac {1}{x}}}

47.

f(x)=\ln x-2e^{x}+{\sqrt {x}}\,

f'(x)=\mathbf {{\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}}

59.

f(x)=\ln(\ln(x^{3}(x+1)))\,

Let

f(x)=g(g(h(x)));\quad g(x)=\ln(x);\quad h(x)=x^{3}(x+1)

. Then

f'(x)={\frac {dg(g(h(x)))}{dg(h(x))}}{\frac {dg(h(x))}{dh(x)}}{\frac {dh(x)}{dx}}={\frac {1}{\ln(x^{3}(x+1))}}{\frac {1}{x^{3}(x+1)}}(4x^{3}+3x^{2})=\mathbf {\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}}

60.

f(x)=\ln(2x^{2}+3x)\,

f'(x)={\frac {1}{2x^{2}+3x}}(4x+3)=\mathbf {\frac {4x+3}{2x^{2}+3x}}

61.

f(x)=\log _{4}x+2\ln x\,

f'(x)=\mathbf {{\frac {1}{x\ln 4}}+{\frac {2}{x}}}

Trigonometric functions

62.

f(x)=3e^{x}-4\cos(x)-{\frac {1}{4}}\ln x\,

f'(x)=\mathbf {3e^{x}+4\sin(x)-{\frac {1}{4x}}}

64.

f(x)=\sin(x)+\cos(x)\,

f'(x)=\mathbf {\cos(x)-\sin(x)}

More Differentiation

65.

{\frac {d}{dx}}[(x^{3}+5)^{10}]

10(x^{3}+5)^{9}(3x^{2})=\mathbf {30x^{2}(x^{3}+5)^{9}}

66.

{\frac {d}{dx}}[x^{3}+3x]

\mathbf {3x^{2}+3}

67.

{\frac {d}{dx}}[(x+4)(x+2)(x-3)]

Let

f(x)=A(x)B(x)C(x);\quad A(x)=x+4;\quad B(x)=x+2;\quad C(x)=x-3

. Then

$f'(x)=A'(x)B(x)C(x)+A(x)B'(x)C(x)+A(x)B(x)C'(x)$
$A'(x)=B'(x)=C'(x)=1$

f'(x)=\mathbf {(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)}

68.

{\frac {d}{dx}}[{\frac {x+1}{3x^{2}}}]

{\frac {3x^{2}-(x+1)(6x)}{(3x^{2})^{2}}}={\frac {3x^{2}-6x^{2}-6x}{9x^{4}}}={\frac {-3x^{2}-6x}{9x^{4}}}=\mathbf {-{\frac {x+2}{3x^{3}}}}

69.

{\frac {d}{dx}}[3x^{3}]

\mathbf {9x^{2}}

70.

{\frac {d}{dx}}[x^{4}\sin x]

\mathbf {4x^{3}\sin x+x^{4}\cos x}

71.

{\frac {d}{dx}}[2^{x}]

\mathbf {\ln(2)2^{x}}

72.

{\frac {d}{dx}}[e^{x^{2}}]

\mathbf {2xe^{x^{2}}}

73.

{\frac {d}{dx}}[e^{2^{x}}]

\mathbf {\ln(2)2^{x}e^{2^{x}}}

Implicit Differentiation

Use implicit differentiation to find y'

74.

x^{3}+y^{3}=xy\,

3x^{2}+3y^{2}y'=y+xy'

$3y^{2}y'-xy'=y-3x^{2}$

\mathbf {y'={\frac {y-3x^{2}}{3y^{2}-x}}}

75.

(2x+y)^{4}+3x^{2}+3y^{2}={\frac {x}{y}}+1\,

4(2x+y)^{3}(2+y')+6x+6yy'={\frac {y-xy'}{y^{2}}}

$8(2x+y)^{3}+4y'(2x+y)^{3}+6x+6yy'={\frac {1}{y}}-{\frac {xy'}{y^{2}}}$
$y'(4(2x+y)^{3}+6y+{\frac {x}{y^{2}}})={\frac {1}{y}}-8(2x+y)^{3}-6x$
$y'({\frac {4y^{2}(2x+y)^{3}+6y^{3}+x}{y^{2}}})={\frac {y-8y^{2}(2x+y)^{3}-6xy^{2}}{y^{2}}}$

\mathbf {y'={\frac {y-8y^{2}(2x+y)^{3}-6xy^{2}}{4y^{2}(2x+y)^{3}+6y^{3}+x}}}

Logarithmic Differentiation

Use logarithmic differentiation to find ${\frac {dy}{dx}}$ :

76.

y=x({\sqrt[{4}]{1-x^{3}}}\,)

\ln y=\ln(x)+\ln({\sqrt[{4}]{1-x^{3}}})=\ln(x)+{\frac {\ln(1-x^{3})}{4}}

${\frac {y'}{y}}={\frac {1}{x}}-{\frac {3x^{2}}{4(1-x^{3})}}$

y'=x({\sqrt[{4}]{1-x^{3}}}\,)({\frac {1}{x}}-{\frac {3x^{2}}{4(1-x^{3})}})=\mathbf {{\sqrt[{4}]{1-x^{3}}}-{\frac {3x^{3}}{4(1-x^{3})^{3/4}}}}

77.

y={\sqrt {x+1 \over 1-x}}\,

\ln y={\frac {1}{2}}(\ln(x+1)-\ln(1-x))

${\frac {y'}{y}}={\frac {1}{2}}({\frac {1}{x+1}}+{\frac {1}{1-x}})$

\mathbf {y'={\frac {1}{2}}{\sqrt {\frac {x+1}{1-x}}}({\frac {1}{x+1}}+{\frac {1}{1-x}})}

78.

y=(2x)^{2x}\,

\ln y=2x\ln(2x)

${\frac {y'}{y}}=2\ln(2x)+2x{\frac {2}{2x}}=2\ln(2x)+2$

\mathbf {y'=(2x)^{2x}(2\ln(2x)+2)}

79.

y=(x^{3}+4x)^{3x+1}\,

\ln y=(3x+1)\ln(x^{3}+4x)

${\frac {y'}{y}}=3\ln(x^{3}+4x)+(3x+1){\frac {3x^{2}+4}{x^{3}+4x}}$

\mathbf {y'=(x^{3}+4x)^{3x+1}(3\ln(x^{3}+4x)+{\frac {(3x+1)(3x^{2}+4)}{x^{3}+4x}})}

80.

y=(6x)^{\cos(x)+1}\,

\ln y=(\cos(x)+1)\ln(6x)

${\frac {y'}{y}}=-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}}$

\mathbf {y'=(6x)^{\cos(x)+1}(-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}})}

Equation of Tangent Line

For each function, $f$ , (a) determine for what values of $x$ the tangent line to $f$ is horizontal and (b) find an equation of the tangent line to $f$ at the given point.

81.

f(x)={\frac {x^{3}}{3}}+x^{2}+5,\;\;\;(3,23)

:

f'(x)=x^{2}+2x

a) $x^{2}+2x=0\implies \mathbf {x=0,-2}$
b) $m=3^{2}+2(3)=9+6=15$

23=15(3)+b\implies b=-22

\mathbf {y=15x-22}

82.

f(x)=x^{3}-3x+1,\;\;\;(1,-1)

:

f'(x)=3x^{2}-3

a) $3x^{2}-3=0\implies \mathbf {x=\pm 1}$
b) $m=3(1)^{2}-3=0$

-1=b

\mathbf {y=-1}

83.

f(x)={\frac {2}{3}}x^{3}+x^{2}-12x+6,\;\;\;(0,6)

:

f'(x)=2x^{2}+2x-12

a) $2x^{2}+2x-12=0=x^{2}+x-6=(x+3)(x-2)\implies \mathbf {x=2,-3}$
b) $m=-12$

6=b

\mathbf {y=-12x+6}

84.

f(x)=2x+{\frac {1}{\sqrt {x}}},\;\;\;(1,3)

:

f'(x)=2-{\frac {1}{2x^{3/2}}}

a) $2-{\frac {1}{2x^{3/2}}}=0\implies 2x^{3/2}={\frac {1}{2}}\implies x^{3/2}={\frac {1}{4}}=2^{-2}\implies \mathbf {x=2^{-4/3}}$
b) $m=2-{\frac {1}{2(1)^{3/2}}}=2-{\frac {1}{2}}={\frac {3}{2}}$

3={\frac {3}{2}}(1)+b\implies b={\frac {3}{2}}

\mathbf {y={\frac {3}{2}}x+{\frac {3}{2}}}

85.

f(x)=(x^{2}+1)(2-x),\;\;\;(2,0)

:

f'(x)=(2x)(2-x)-(x^{2}+1)=4x-2x^{2}-x^{2}-1=-3x^{2}+4x-1

a) $-3x^{2}+4x-1=0=(3x-1)(-x+1)\implies \mathbf {x=1,{\frac {1}{3}}}$
b) $m=-3(2)^{2}+4(2)-1=-3(4)+8-1=-12+7=-5$

0=-5(2)+b\implies b=10

\mathbf {y=-5x+10}

86.

f(x)={\frac {2}{3}}x^{3}+{\frac {5}{2}}x^{2}+2x+1,\;\;\;(3,{\frac {95}{2}})

:

f'(x)=2x^{2}+5x+2

a) $2x^{2}+5x+2=0=(2x+1)(x+2)\implies \mathbf {x=-{\frac {1}{2}},-2}$
/ b) $m=2(3)^{2}+5(3)+2=18+15+2=35$

{\frac {95}{2}}=35(3)+b\implies b=-{\frac {115}{2}}

\mathbf {y=35x-{\frac {115}{2}}}

87. Find an equation of the tangent line to the graph defined by

(x-y-1)^{3}=x\,

at the point (1,-1).

3(x-y-1)^{2}(1-y')=1

$1-y'={\frac {1}{3(x-y-1)^{2}}}$
$y'=1-{\frac {1}{3(x-y-1)^{2}}}$
$m=1-{\frac {1}{3(1-(-1)-1)^{2}}}=1-{\frac {1}{3(1)^{2}}}={\frac {2}{3}}$
$-1={\frac {2}{3}}(1)+b\implies b=-{\frac {5}{3}}$

\mathbf {y={\frac {2}{3}}x-{\frac {5}{3}}}

88. Find an equation of the tangent line to the graph defined by

e^{xy}+x^{2}=y^{2}\,

at the point (1,0).

y'e^{xy}+2x=2yy'

$y'(e^{xy}-2y)=-2x$
$y'={\frac {2x}{2y-e^{xy}}}$
$m={\frac {2(1)}{2(0)-e^{1(0)}}}={\frac {2}{-1}}=-2$
$0=-2(1)+b\implies b=2$

\mathbf {y=-2x+2}

Higher Order Derivatives

89. What is the second derivative of

3x^{4}+3x^{2}+2x

?

{\frac {d}{dx}}(3x^{4}+3x^{2}+2x)=12x^{3}+6x+2

{\frac {d^{2}}{dx^{2}}}(3x^{4}+3x^{2}+2x)={\frac {d}{dx}}(12x^{3}+6x+2)=\mathbf {36x^{2}+6}

90. Use induction to prove that the (n+1)th derivative of a n-th order polynomial is 0.

base case: Consider the zeroth-order polynomial,

c

.

{\frac {dc}{dx}}=0

induction step: Suppose that the n-th derivative of a (n-1)th order polynomial is 0. Consider the n-th order polynomial, $f(x)$ . We can write $f(x)=cx^{n}+P(x)$ where $P(x)$ is a (n-1)th polynomial.

{\frac {d^{n+1}}{dx^{n+1}}}f(x)={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n}+P(x))={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n})+{\frac {d^{n+1}}{dx^{n+1}}}P(x)={\frac {d^{n}}{dx^{n}}}(cnx^{n-1})+{\frac {d}{dx}}{\frac {d^{n}}{dx^{n}}}P(x)=0+{\frac {d}{dx}}0=0

Advanced Understanding of Derivatives

91. Let

f^{\prime }(x)

be the derivative of

f(x)

. Prove the derivative of

-f(x)

is

-f^{\prime }(x)

.

Suppose

f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}

. Let

g(x)=-f(x)

.

${\begin{aligned}g^{\prime }(x)&=\lim _{h\to 0}{\dfrac {g(x+h)-g(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-f(x+h)+f(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-\left(f(x+h)-f(h)\right)}{h}}\\&=-\lim _{h\to 0}{\dfrac {\left(f(x+h)-f(h)\right)}{h}}\\&=-f^{\prime }(x)\end{aligned}}$