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Calculus 2 : Definition of Derivative

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Example Question #51 : Derivatives

Find the derivative of f(x)=sinx using the definition of the derivative.

Possible Answers:

f'(x)=-cos(x)

f'(x)=-sin(x)

f'(x)=cos(x)

f'(x)=sin(x)

Correct answer:

f'(x)=cos(x)

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

f(x)=sinx

f(x+h)=sin(x+h)

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{sin(x+h)-sin(x)}{h}$

Substituting in the trigonometric identity

sin(A+B)=sinAcosB+cosAsinB

$f'(x)=\lim_{h\rightarrow 0}\frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{sin(x)(cos(h)-1)+cos(x)sin(h)}{h}$

Then using the identities

$\lim_{h\rightarrow 0}\frac{cos(h)-1}{h}=0$ and $\lim_{h\rightarrow 0}\frac{sin(h)}{h}=1$

$f'(x)=\lim_{h\rightarrow 0}\left [sin(x)\frac{(cos(h)-1)}{h}+cos(x)\frac{sin(h)}{h} \right ]$

$f'(x)=sin(x)\lim_{h\rightarrow 0}\left [\frac{(cos(h)-1)}{h} \right ]+cos(x)\lim_{h\rightarrow 0}\left [\frac{sin(h)}{h} \right ]$

f'(x)=sin(x)*0+cos(x)*1

f'(x)=cos(x)

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Example Question #52 : Derivatives

Find the derivative of f(x)=cosx using the definition of the derivative.

Possible Answers:

f'(x)=sin(x)

f'(x)=cos(x)

f'(x)=-sin(x)

f'(x)=-cos(x)

Correct answer:

f'(x)=sin(x)

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

f(x)=cosx

f(x+h)=cos(x+h)

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{cos(x+h)-cos(x)}{h}$

Substituting in the trigonometric identity

cos(A-B)=cosAcosB+sinAsinB

$f'(x)=\lim_{h\rightarrow 0}\frac{cos(x)cos(h)+sin(x)sin(h)-cos(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{cos(x)(cos(h)-1)+sin(x)sin(h)}{h}$

Then using the identities

$\lim_{h\rightarrow 0}\frac{cos(h)-1}{h}=0$ and $\lim_{h\rightarrow 0}\frac{sin(h)}{h}=1$

$f'(x)=\lim_{h\rightarrow 0}\left [cos(x)\frac{(cos(h)-1)}{h}+sin(x)\frac{sin(h)}{h} \right ]$

$f'(x)=cos(x)\lim_{h\rightarrow 0}\left [\frac{(cos(h)-1)}{h} \right ]+sin(x)\lim_{h\rightarrow 0}\left [\frac{sin(h)}{h} \right ]$

f'(x)=cos(x)*0+sin(x)*1

f'(x)=sin(x)

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Example Question #53 : Derivatives

Find the derivative of $f(x)=\frac{x+3}{x-2}$ using the definition of the derivative.

Possible Answers:

f'(x)=1

$f'(x)=\frac{-5}{x^{2}-4x+4}$

$f'(x)=\frac{x+3}{x-2}$

$f'(x)=\frac{3}{-2}$

Correct answer:

$f'(x)=\frac{-5}{x^{2}-4x+4}$

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=\frac{x+3}{x-2}$

$f(x+h)=\frac{x+h+3}{x+h-2}$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{x+h+3}{x+h-2}-\frac{x+3}{x-2}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{x+h+3}{x+h-2}-\frac{x+3}{x-2} \right )$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{(x+h+3)(x-2)-(x+3)(x+h-2)}{(x+h-2)(x-2)}\right )$

$f'(x)=\lim_{h\rightarrow 0} \frac{(x+h+3)(x-2)-(x+3)(x+h-2)}{h(x+h-2)(x-2)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(x^{2}+hx+3x-2x-2h-6)-(x^{2}+hx-2x+3x+3h-6)}{h(x^{2}+hx-2x-2x-2h+4)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{x^{2}+hx+x-2h-6-x^{2}-hx-x-3h+6}{h(x^{2}+hx-4x-2h+4)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{-5h}{h(x^{2}+hx-4x-2h+4)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{-5}{(x^{2}+hx-4x-2h+4)}$

$f'(x)=\frac{-5}{x^{2}+(0)x-4x-2(0)+4}$

$f'(x)=\frac{-5}{x^{2}-4x+4}$

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Example Question #54 : Derivatives

Find the derivative of $f(x)=\frac{5x^2-1}{x+3}$ using the definition of the derivative.

Possible Answers:

f'(x)={10x}

$f'(x)=\frac{x^2+6x-2}{x^2+3x-7}$

$f'(x)=\frac{5x^2+30x+1}{x^2+6x+9}$

$f'(x)=\frac{5x^2-1}{x+3}$

Correct answer:

$f'(x)=\frac{5x^2+30x+1}{x^2+6x+9}$

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

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

$f(x+h)=\frac{5(x+h)^2-1}{x+h+3}$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{5(x+h)^2-1}{x+h+3}-\frac{5(x)^2-1}{x+3}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{5(x+h)^2-1}{x+h+3}-\frac{5(x)^2-1}{x+3} \right )$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5(x+h)^{2}-1)(x+3)-(5x^{2}-1)(x+h+3)}{h(x+h+3)(x+3)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5(x^2+2xh+h^2)-1)(x+3)-(5x^{2}-1)(x+h+3)}{h(x+h+3)(x+3)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5x^2+10xh+5h^2-1)(x+3)-(5x^{2}-1)(x+h+3)}{h(x+h+3)(x+3)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5x^3+10x^2h+5h^2x-x+15x^2+30xh+15h^2-3)-(5x^3+5hx^2+15x^2-x-h-3)}{h(x^2+hx+3x+3x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{5x^3+10x^2h+5h^2x-x+15x^2+30xh+15h^2-3-5x^3-5hx^2-15x^2+x+h+3}{h(x^2+hx+3x+3x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{5x^2h+5h^2x+30xh+15h^2+h}{h(x^2+hx+6x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{h(5x^2+5hx+30x+15h+1)}{h(x^2+hx+6x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{5x^2+5hx+30x+15h+1}{x^2+hx+6x+3h+9}$

$f'(x)=\frac{5x^2+5(0)x+30x+15(0)+1}{x^2+(0)x+6x+3(0)+9}$

$f'(x)=\frac{5x^2+30x+1}{x^2+6x+9}$

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Example Question #55 : Derivatives

Find the derivative of $f(x)=x^{3}-3x^{2}-6x+10$ using the definition of the derivative.

Possible Answers:

f'(x)=3x^2-6x-6

f(x)=10

$f'(x)=x^{3}-3x^{2}-6x+10$

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

Correct answer:

f'(x)=3x^2-6x-6

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=x^{3}-3x^{2}-6x+10$

$f(x+h)=(x+h)^{3}-3(x+h)^{2}-6(x+h)+10$

f(x+h)=(x^3+3xh^2+3x^2h+h^3)-3(x^2+2xh+h^2)-6(x+h)+10

f(x+h)=x^3+3xh^2+3x^2h+h^3-3x^2-6xh-3h^2-6x-6h+10

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{x^3+3xh^2+3x^2h+h^3-3x^2-6xh-3h^2-6x-6h+10-(x^{3}-3x^{2}-6x+10)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^3+3xh^2+3x^2h+h^3-3x^2-6xh-3h^2-6x-6h+10-x^{3}+3x^{2}+6x-10}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{3xh^2+3x^2h+h^3-6xh-3h^2-6h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h(3xh+3x^2+h^2-6x-3h-6)}{h}$

$f'(x)=\lim_{h\rightarrow 0}3xh+3x^2+h^2-6x-3h-6$

f'(x)=3x(0)+3x^2+(0)^2-6x-3(0)-6

f'(x)=3x^2-6x-6

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Example Question #56 : Derivatives

Find the derivative of $f(x)=\sqrt{x^{2}-4x}$ using the definition of the derivative.

Possible Answers:

$f'(x)=\sqrt{2x-4}$

$f'(x)=\sqrt{x^{2}-4x}$

$f'(x)=\frac{x-2}{\sqrt{x^2-4x}}$

$f'(x)=\frac{2x-4}{\sqrt{x^2-4x}}$

Correct answer:

$f'(x)=\frac{x-2}{\sqrt{x^2-4x}}$

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=\sqrt{x^{2}-4x}$

$f(x+h)=\sqrt{(x+h)^{2}-4(x+h)}$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{(x+h)^{2}-4(x+h)}-\sqrt{x^{2}-4x}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{x^2+2xh+h^2-4x-4h}-\sqrt{x^{2}-4x}}{h}$

Next we will multiply the numerator and denominator by the complex conjugate of the numerator

$f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{x^2+2xh+h^2-4x-4h}-\sqrt{x^{2}-4x}}{h}*\frac{\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x}}{\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x}}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\left (\sqrt{x^2+2xh+h^2-4x-4h} \right )^2-\left (\sqrt{x^{2}-4x} \right )^2}{h*(\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{ x^2+2xh+h^2-4x-4h-(x^{2}-4x)}{h*(\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{ 2xh+h^2-4h}{h*(\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h*( 2x+h-4)}{h*(\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{( 2x+h-4)}{\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x}}$

$f'(x)=\frac{2x+(0)-4}{\sqrt{x^2+2x(0)+(0)^2-4x-4(0)}+\sqrt{x^{2}-4x}}$

$f'(x)=\frac{2x-4}{\sqrt{x^2-4x}+\sqrt{x^{2}-4x}}$

$f'(x)=\frac{2(x-2)}{2\sqrt{x^2-4x}}$

$f'(x)=\frac{x-2}{\sqrt{x^2-4x}}$

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Example Question #57 : Derivatives

Find the derivative of f(x)=(x-7)(x-9) using the definition of the derivative.

Possible Answers:

f'(x)=7

f'(x)=x^2

f'(x)=2x-16

f'(x)=5x-2

Correct answer:

f'(x)=2x-16

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

f(x)=(x-7)(x-9)

f(x)=x^2-7x-9x+63

f(x)=x^2-16x+63

f(x+h)=(x+h-7)(x+h-9)

f(x+h)=(x^2+hx-7x+hx+h^2-7h-9x-9h+63)

f(x+h)=(x^2+2hx-16x+h^2-16h+63)

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{(x^2+2hx-16x+h^2-16h+63)-(x^2-16x+63)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^2+2hx-16x+h^2-16h+63-x^2+16x-63}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2hx+h^2-16h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h*(2x+h-16)}{h}$

$f'(x)=\lim_{h\rightarrow 0}(2x+h-16)$

f'(x)=2x+0-16

f'(x)=2x-16

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Example Question #58 : Derivatives

Find the derivative of $f(x)=\sqrt{x-7}$ using the definition of the derivative.

Possible Answers:

$f'(x)=\frac{1}{2\sqrt{x-7}}$

f(x)=1

$f'(x)=\sqrt{x-7}$

$f'(x)=\frac{x-7}{\sqrt{x-7}}$

Correct answer:

$f'(x)=\frac{1}{2\sqrt{x-7}}$

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=\sqrt{x-7}$

$f(x+h)=\sqrt{x+h-7}$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{x+h-7}-\sqrt{x-7}}{h}$

Next we will multiply the numerator and denominator by the complex conjugate of the numerator

$f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{x+h-7}-\sqrt{x-7}}{h}*\frac{\sqrt{x+h-7}+\sqrt{x-7}}{\sqrt{x+h-7}+\sqrt{x-7}}$

$f'(x)=\lim_{h\rightarrow 0}\frac{(\sqrt{x+h-7})^2-(\sqrt{x-7})^2}{h(\sqrt{x+h-7}+\sqrt{x-7})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x+h-7-(x-7)}{h(\sqrt{x+h-7}+\sqrt{x-7})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x+h-7-x+7}{h(\sqrt{x+h-7}+\sqrt{x-7})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h}{h(\sqrt{x+h-7}+\sqrt{x-7})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{\sqrt{x+h-7}+\sqrt{x-7}}$

$f'(x)=\frac{1}{\sqrt{x+(0)-7}+\sqrt{x-7}}$

$f'(x)=\frac{1}{2\sqrt{x-7}}$

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Example Question #59 : Derivatives

Find the derivative of f(x)=(2x+3)(x-16) using the definition of the derivative.

Possible Answers:

f'(x)=1

f'(x)=3x-16

f'(x)=2x^2

f'(x)=4x-29

Correct answer:

f'(x)=4x-29

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

f(x)=(2x+3)(x-16)

f(x)=2x^2+3x-32x-48

f(x)=2x^2-29x-48

f(x+h)=(2(x+h)+3)(x+h-16)

f(x+h)=(2x+2h+3)(x+h-16)

f(x+h)=(2x^2+2hx+3x+2xh+2h^2+3h-32x-32h-48)

f(x+h)=2x^2+2h^2+4hx-29x-29h-48

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{(2x^2+2h^2+4hx-29x-29h-48)-(2x^2-29x-48)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2x^2+2h^2+4hx-29x-29h-48-2x^2+29x+48}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2h^2+4hx-29h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h(2h+4x-29)}{h}$

$f'(x)=\lim_{h\rightarrow 0}(2h+4x-29)$

f'(x)=(2(0)+4x-29)

f'(x)=4x-29

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Example Question #51 : Derivative Review

Find the derivative of f(x)=(x+2)(x+3)(x+4) using the definition of the derivative.

Possible Answers:

f'(x)=5x^2-x +3

f'(x)=-x^2+7x +37

f'(x)=4x^2+7x +49

f'(x)=3x^2+18x +26

Correct answer:

f'(x)=3x^2+18x +26

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

f(x)=(x+2)(x+3)(x+4)

f(x)=(x^2+5x+6)(x+4)

f(x)=x^3+5x^2+6x +4x^2+20x+24

f(x)=x^3+9x^2+26x+24

f(x+h)=(x+h+2)(x+h+3)(x+h+4)

f(x+h)=(x^2+hx+2x +hx+h^2+2h +3x+3h+6)(x+h+4)

f(x+h)=(x^2+2hx+5x+h^2+5h+6)(x+h+4)

f(x+h)=(x^3+2hx^2+5x^2+h^2x+5hx+6x +x^2h+2h^2x+5xh+h^3+5h^2+6h +4x^2+8hx+20x+4h^2+20h+24)

f(x+h)=(x^3+h^3+3x^2h+3h^2x+9x^2+9h^2+18hx +26x+26h +24)

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{(x^3+h^3+3x^2h+3h^2x+9x^2+9h^2+18hx +26x+26h +24)-(x^3+9x^2+26x+24)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^3+h^3+3x^2h+3h^2x+9x^2+9h^2+18hx +26x+26h +24-x^3-9x^2-26x-24}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h^3+3x^2h+3h^2x+9h^2+18hx +26h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h*(h^2+3x^2+3hx+9h+18x +26)}{h}$

$f'(x)=\lim_{h\rightarrow 0}(h^2+3x^2+3hx+9h+18x +26)$

f'(x)=(0)^2+3x^2+3(0)x+9(0)+18x +26

f'(x)=3x^2+18x +26

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Calculus 2 : Definition of Derivative

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Example Question #52 : Derivatives

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