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Calculus 2 : First and Second Derivatives of Functions

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Example Question #1 : First And Second Derivatives Of Functions

Find the second derivative of the following equation:

f(x)=ln(2x-1)

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find the second derivative, first we need to find the first derivative. The derivative of a natural log is the derivative of operand times the inverse of the operand. So for the given function, we get the first derivative to be

$f'(x)=2\cdot \frac{1}{2x-1}=\frac{2}{2x-1}$ .

Now, we have to take the derivative of the first derivative. To simplify this, we can rewrite the function to be f'(x)=2(2x-1)^-^1 . From here we can use the chain rule to solve for the derivative. First, multiply by the exponent and find the new exponent by subtracting the old one by one. Next multiply by the derivative of (2x-1) and then simplify. Thus, we get

$f''(x)=2\cdot-1\cdot2(2x-1)^-^2=-\frac{4}{(2x-1)^2}$ .

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Example Question #2 : First And Second Derivatives Of Functions

Find the second derivative of the following function.

f(x)=sec(x)

Possible Answers:

f''(x)=-csc(x)

f''(x)=cot(x)csc(x)

$f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))$

f''(x)=sec^3(x)tan(x)

f''(x)=sec(x)tan(x)

Correct answer:

$f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))$

Explanation:

To find the second derivative, first we need to find the first derivative. So for the given function, we get the first derivative to be

f'(x)=tan(x)sec(x) .

Now we have to take the derivative of the derivative. To do this we need to use the product rule as shown below

Thus, we get

$\\f''(x)=tan(x)\cdot \frac{\mathrm{d} }{\mathrm{d} x}sec(x)+sec(x)\cdot \frac{\mathrm{d} }{\mathrm{d} x}tan(x)\\f''(x)=tan(x)\cdot sec(x)\cdot tan(x)+sec(x)\cdot sec^2(x)\\f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))$ .

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Example Question #1 : First And Second Derivatives Of Functions

Find the second derivative of the given function:

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

Possible Answers:

x^2+2x+1

$\frac{x^2-1}{x^2}$

$\frac{2}{x^3}$

$\frac{2x+2}{x}$

Correct answer:

$\frac{2}{x^3}$

Explanation:

To find the second derivative, first we need to find the first derivative. To find the first derivative we need to use the quotient rule as follows. So for the given function, we get the first derivative to be
$\\f'(x)=\frac{x\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2x+1)-(x^2+2x+1)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x)}{x^2}\\f'(x)=\frac{x\cdot (2x+2)-(x^2+2x+1)\cdot 1}{x^2}\\f'(x)=\frac{x^2-1}{x^2}$ .

Now we have to take the derivative of the derivative. To do this we need to use the quotient rule as shown below.

Thus, we get

$\\f''(x)=\frac{x^2\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2-1)-(x^2-1)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2)}{x^4}\\ f''(x)=\frac{2x^3-2x^3+2x}{x^4}\\f''(x)=\frac{2x}{x^4}=\frac{2}{x^3}$

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Example Question #1 : First And Second Derivatives Of Functions

f(x)= 2sin(2x)+e^x

Calculate f''(x).

Possible Answers:

$f''(x)=-4cos(2x)+e^{2x}$

$f''(x)=4sin(x)+e^{2x}$

f''(x)=-8cos(2x)+e^x

f''(x)= 8sin(2x)+e^x

f''(x)= -8sin(2x)+e^x

Correct answer:

f''(x)= -8sin(2x)+e^x

Explanation:

There are two seprate functions that make up f(x) . There is 2sin(2x) and e^x .

On a general note,

$\frac{d}{dx}sin(ax)=acos(ax)$ and

$\frac{d}{dx}cos(ax)=-asin(ax)$ .

Also,

$\frac{d}{dx}e^u=e^u \cdot u'$

With that said, let's calculate f'(x) :

f'(x)= 4cos(2x)+e^x . Notice the e^x term is still unchanged.

And now let's calculate f''(x) .

f'(x)= -8sin(2x)+e^x

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Example Question #5 : First And Second Derivatives Of Functions

$f(x)= \frac{1}{2}e^{2x}+\frac{1}{8}x^2$

$f'(x)=ae^{bx} + \frac{1}{4}x$

Find and .

Possible Answers:

a=2, b=2

$a=\frac{1}{2}$ , b=2

$a=\frac{1}{2}$ , b=1

a=0, b=1

a=1, b=2

Correct answer:

a=1, b=2

Explanation:

To find a and b, first let's calculate f'(x) .

$f'(x)=\frac{1}{2}(2)e^{2x}+\frac{1}4{x}$

Remember that $\frac{d}{dx}e^{ax}=ae^{ax}$ , where a is real number.

is simply the coefficient in front of the exponential, which simplifies to 1, and is the power of the exponent, which is 2.

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Example Question #1 : First And Second Derivatives Of Functions

Determine the derivative of y^2=x^2+5y with the respect to .

Possible Answers:

$\frac{dy}{dx}=2y=2x+5$

$\frac{dy}{dx}=\frac{2x+5}{2y}$

$\frac{dy}{dx}=\frac{2x+5}{2}$

$\frac{dy}{dx}=\frac{2x}{2y-5}$

$\frac{dy}{dx}=\frac{2x-5}{2y}$

Correct answer:

$\frac{dy}{dx}=\frac{2x}{2y-5}$

Explanation:

In order so solve the derivative with the respect to x, implicit differentiation is required. The notation for finding the derivative of the function with the respect to x is:

$\frac{dy}{dx}$

Take the derivative and apply chain rule where necessary.

y^2=x^2+5y

$2y \cdot \frac{dy}{dx}=2x +5 \cdot \frac{dy}{dx}$

$2y \cdot \frac{dy}{dx}-5 \cdot \frac{dy}{dx}=2x$

$\frac{dy}{dx}[2y-5]=2x$

$\frac{dy}{dx}=\frac{2x}{2y-5}$

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Example Question #7 : First And Second Derivatives Of Functions

Find the derivative of $\small lnx^{lnx}$ .

Possible Answers:

$\small \small lnx^{lnx}(ln(lnx)+1/x)$

$\small (lnxlnx^{lnx-1})/x$

$\small lnxlnx/x$

$\small \small lnx^{lnx}(ln(lnx)/x+1/x)$

$\small \small lnx^{lnx}ln(lnx)/x$

Correct answer:

$\small \small lnx^{lnx}(ln(lnx)/x+1/x)$

Explanation:

To solve this derivative, we need to use logarithmic differentiation. This allows us to use the logarithm rule $\small lna^{b}=blna$ to solve an easier derivative.

Let $\small y=lnx^{lnx}$ .

Now we'll take the natural log of both sides to get

$\small lny=ln(lnx^{lnx})=lnxln(lnx)$ .

Now we can use implicit differentiation to solve for $\small y'$ .

The derivative of $\small lny$ is $\small y'/y$ , and the derivative of $\small \small \small lnx(ln(lnx))$ can be found using the product rule, which states

$\small \frac{\mathrm{d} }{\mathrm{d} x}(u\cdot v)=u'v+uv'$ where $\small u$ and $\small v$ are functions of $\small x$ .

Letting $\small u=lnx$ and $\small v=ln(lnx)$

(which means $\small u'=1/x$ and $\small v'=1/(xlnx)$ ) we get our derivative to be $\small ln(lnx)/x+1/x$ .

Now we have $\small y'/y=ln(lnx)/x+1/x$ , but $\small y=lnx^{lnx}$ , so subbing that in we get

$\small \small y'/(lnx^{lnx})=ln(lnx)/x+1/x$ .

Multiplying both sides by $\small lnx^{lnx}$ , we get

$\small y'=lnx^{lnx}(ln(lnx)/x+1/x)$ .

That is our derivative.

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Example Question #8 : First And Second Derivatives Of Functions

Find the first derivative of the function:

$f(x)=\cos(x^3)+\sqrt{x^2+1}$

Possible Answers:

$-3x^2\sin(x^3)+\frac{x}{\sqrt{x^2+1}}$

$-\sin(x^3)+\frac{x}{\sqrt{x^2+1}}$

$3x^2\sin(x^3)+\frac{x}{\sqrt{x^2+1}}$

$-3x^2\sin(x^3)+{x}{\sqrt{x^2+1}}$

Correct answer:

$-3x^2\sin(x^3)+\frac{x}{\sqrt{x^2+1}}$

Explanation:

The derivative of the function is equal to

$-3x^2\sin(x^3)+\frac{x}{\sqrt{x^2+1}}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$

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Example Question #9 : First And Second Derivatives Of Functions

Find the second derivative of the following function:

$f(x)=2xe^{3x}+\tan(x)$

Possible Answers:

$12e^{3x}+18xe^{3x}+2\sec(x)\tan(x)$

$6e^{3x}+18xe^{3x}+2\sec^2(x)\tan(x)$

$18xe^{3x}+2\sec^2(x)\tan(x)$

$12e^{3x}+18xe^{3x}+2\sec^2(x)\tan(x)$

Correct answer:

$12e^{3x}+18xe^{3x}+2\sec^2(x)\tan(x)$

Explanation:

The first derivative of the function is equal to

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

The second derivative of the function (the derivative of the above function) is

$f''(x)=12e^{3x}+18xe^{3x}+2\sec^2(x)\tan(x)$

The following rules were used for the derivatives:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$

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Example Question #1 : Velocity, Speed, Acceleration

The position of a car is given by the following function:

$f(x)=\cos(x)e^{11x}+\csc(x)$

What is the velocity function of the car?

Possible Answers:

$-\sin(x)e^{11x}+11\cos(x)e^{11x}$

$-\sin(x)e^{11x}+\cos(x)e^{11x}-\csc(x)\cot(x)$

$\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}+\csc(x)\cot(x)$

Correct answer:

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

Explanation:

The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

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Calculus 2 : First and Second Derivatives of Functions

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #1 : First And Second Derivatives Of Functions

Example Question #2 : First And Second Derivatives Of Functions

Example Question #1 : First And Second Derivatives Of Functions

Example Question #1 : First And Second Derivatives Of Functions

Example Question #5 : First And Second Derivatives Of Functions

Example Question #1 : First And Second Derivatives Of Functions

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