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Calculus 2 : Definition of Derivative

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Example Question #61 : Definition Of Derivative

Find the derivative of $f(x)=\frac{x}{x+1}$ using the definition of the derivative.

Possible Answers:

f'(x)=1

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

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

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

Correct answer:

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

Explanation:

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

$f(x)=\frac{x}{x+1}$

$f(x+h)=\frac{x+h}{x+h+1}$

Substituting these expressions into the definition of a derivative gives us

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

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

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

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

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

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

$f'(x)=\left (\frac{1}{x^2+(0)x+2x+(0)+1} \right )$

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

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Example Question #61 : Definition Of Derivative

Possible Answers:

4x+C

x^2

Correct answer:

Explanation:

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Example Question #61 : Definition Of Derivative

What is the derivative of a term: ax^2 ?

Possible Answers:

3x^2

2ax

2x^3

Correct answer:

2ax

Explanation:

Step 1: Move the exponent to the coefficient..

We get, 2ax..

Step 2: The new exponent after moving the old one down is less than it was before:

$2ax^{2-1}=2ax$

The derivative of ax^2 is 2ax

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Example Question #64 : Definition Of Derivative

Find the function represented by the following limit,

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}$

Possible Answers:

Constant

$\frac{1}{y^2}+1$

$\frac{2}{y}$

y^2 -1

$\ln(y+4)$

Correct answer:

$\frac{2}{y}$

Explanation:

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}$ (1)

Recall the first principles definition of a derivative,

$f'(y) = \lim_{h->0}\frac{f(y+h)-f(h)}{h}$ (2)

The limit in Equation (1) can be determined by inspection if we compare to the general definition of a derivative, Equation (2). It's apparent that equation (1) is the definition of the derivative for the function $\ln(y^2)$ . So the limit can be found by simply differentiating the function $\ln(y^2)$ with respect to .

$\frac{d}{dy}(\ln(y^2))=(2y)\frac{1}{y^2}=\frac{2}{y}$

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}=\frac{2}{y}$

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Example Question #65 : Definition Of Derivative

Find the derivative of the following function:

$f(x)=\frac{cosx}{sinx}$ .

Possible Answers:

secxtanx

-csc^2x

-sec^2x

sec^2x

csc^2x

Correct answer:

-csc^2x

Explanation:

The function in the problem is simply cotx ! Therefore, by knowing the derivative of cotx , we know the solution is -csc^2x . You would obtain the same results if you completed this problem using the quotient rule and simplifying.

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Calculus 2 : Definition of Derivative

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #61 : Definition Of Derivative

Example Question #61 : Definition Of Derivative

Example Question #61 : Definition Of Derivative

Example Question #64 : Definition Of Derivative

Example Question #65 : Definition Of Derivative

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