High School Math : Derivatives

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Finding Second Derivative Of A Function

What is the second derivative of \(\displaystyle (x+6)^2\)?

Possible Answers:

\(\displaystyle \frac{1}{2}\)

\(\displaystyle 2x\)

\(\displaystyle 6x+12\)

\(\displaystyle 2\)

\(\displaystyle 2x+12\)

Correct answer:

\(\displaystyle 2\)

Explanation:

To find the second derivative, we need to start with the first derivative.

To solve for the first derivative, we're going to use the chain rule. The chain rule says that when taking the derivative of a nested function, your answer is the derivative of the outside times the derivative of the inside.

Mathematically, it would look like this: \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}f(g(x))=f'(g(x))*g'(x)\)

Plug in our equations.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=(2(x+6)^{1})*(1+0)\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=(2(x+6))\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=2x+12\)

From here, we can use our normal power rule to find the second derivative.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=(1*2x^{1-1})+(0*12x^{0-1})\)

Anything times zero is zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=(1*2x^{1-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=(2x^{0})\)

Anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=2\)

Example Question #42 : Derivatives

What is the second derivative of \(\displaystyle 4x^2+5x-3\)?

Possible Answers:

\(\displaystyle y=8\)

\(\displaystyle y=3\)

\(\displaystyle y=32x+5x^2\)

\(\displaystyle y=16x\)

\(\displaystyle y=8x\)

Correct answer:

\(\displaystyle y=8\)

Explanation:

To find the second derivative, we need to find the first derivative first. To find the first derivative, we can use the power rule.

For each variable, multiply by the exponent and reduce the exponent by one:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=(2*4x^{2-1})+(1*5x^{1-1})-(0*3x^0-1)\)

Treat \(\displaystyle 3\) as \(\displaystyle 3x^0\) since anything to the zero power is one.

Remember, anything times zero is zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=(8x^{1})+(5x^{0})-0\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=8x+5\)

Now follow the same process but for \(\displaystyle 8x+5\).

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=(1*8x^{1-1})+(0*5x^{0-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=(1*8x^{0})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=8\)

Therefore the second derivative will be the line \(\displaystyle y=8\).

Example Question #1 : Finding Second Derivative Of A Function

Let \(\displaystyle f(x)=sin(x)-cos(x)\).

Find the second derivative of \(\displaystyle f(x)\).

Possible Answers:

\(\displaystyle f''(x)=sin^2(x)-cos^2(x)\)

\(\displaystyle f''(x)=sin^2(x)+cos^2(x)\)

\(\displaystyle f''(x)=sin(x)+cos(x)\)

\(\displaystyle f''(x)=sin(x)-cos(x)\)

\(\displaystyle f''(x)=cos(x)-sin(x)\)

Correct answer:

\(\displaystyle f''(x)=cos(x)-sin(x)\)

Explanation:

The second derivative is just the derivative of the first derivative. So first we find the first derivative of \(\displaystyle f(x)\). Remember the derivative of \(\displaystyle \sin x\) is \(\displaystyle \cos x\), and the derivative for \(\displaystyle \cos x\) is \(\displaystyle -\sin x\).

 \(\displaystyle f'(x)= cos(x)+sin(x)\)

 

Then to get the second derivative, we just derive this function again. So

\(\displaystyle f''(x)=-sin(x)+cos(x)=cos(x)-sin(x)\)

Example Question #41 : Derivatives

Define \(\displaystyle f (x) = 6x^{3} - 12x^{2} + 4x -8\).

What is \(\displaystyle f ' ' (x)\)?

Possible Answers:

\(\displaystyle f ' '(x) = 6x - 1 2\)

\(\displaystyle f ' '(x) = 36x - 24\)

\(\displaystyle f ' '(x) = 36x\)

\(\displaystyle f ' '(x) = 18\)

\(\displaystyle f ''(x) = 18x - 24\)

Correct answer:

\(\displaystyle f ' '(x) = 36x - 24\)

Explanation:

Take the derivative \(\displaystyle f'\) of \(\displaystyle f\), then take the derivative of \(\displaystyle f'\).

\(\displaystyle f (x) = 6x^{3} - 12x^{2} + 4x -8\)

 

\(\displaystyle f '(x) = 3 \cdot 6x^{3-1} - 2 \cdot 12x^{2-1} + 4 -0\)

\(\displaystyle f '(x) = 18x^{2} - 24x+ 4\)

 

\(\displaystyle f ' '(x) = 2 \cdot 18x^{2-1} - 24+ 0\)

\(\displaystyle f ' '(x) = 36x - 24\)

Example Question #302 : Computation Of The Derivative

Define \(\displaystyle g(x) = 7x^{5} - 6x^{3} - 17x\).

What is \(\displaystyle g ''(x)\) ?

Possible Answers:

\(\displaystyle g'' (x) = 140x^{3} - 18x\)

\(\displaystyle g'' (x) = 140x^{3} - 36x\)

\(\displaystyle g'' (x) = 140x^{3} - 36x -17\)

\(\displaystyle g'' (x) = 70x^{3} - 18x\)

\(\displaystyle g'' (x) = 70x^{3} - 36x\)

Correct answer:

\(\displaystyle g'' (x) = 140x^{3} - 36x\)

Explanation:

Take the derivative \(\displaystyle g'\) of \(\displaystyle g\), then take the derivative of \(\displaystyle g'\).

\(\displaystyle g(x) = 7x^{5} - 6x^{3} - 17x\)

 

\(\displaystyle g' (x) = 5\cdot 7x^{5-1} - 3\cdot 6x^{3-1} - 17\)

\(\displaystyle g' (x) = 35x^{4} - 18x^{2} - 17\)

 

\(\displaystyle g'' (x) = 4\cdot 35x^{4-1} - 2\cdot 18x^{2-1} - 0\)

\(\displaystyle g'' (x) = 140x^{3} - 36x\)

Example Question #1211 : Ap Calculus Ab

Define \(\displaystyle g(x) = \cos \left (x^{2} \right )\).

What is \(\displaystyle g''(x)\)?

Possible Answers:

\(\displaystyle g''(x) =-4x^2 \cos(x^2)\)

\(\displaystyle g''(x) = -4x^2 \sin(x^2)\)

\(\displaystyle g''(x) = - 2 \cos(x^2)-4x^2 \sin(x^2) \right ]\)

\(\displaystyle g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]\)

\(\displaystyle g''(x) =4x^2 \cos(x^2)\)

Correct answer:

\(\displaystyle g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]\)

Explanation:

Take the derivative \(\displaystyle g'\) of \(\displaystyle g\), then take the derivative of \(\displaystyle g'\).

\(\displaystyle g(x) = \cos \left (x^{2} \right )\)

 

\(\displaystyle g'(x) = 2x \cdot [-\sin (x^2)]\)

\(\displaystyle g'(x) = - 2x \sin (x^2)\)

 

\(\displaystyle g''(x) = - 2\left [ 1 \cdot \sin(x^2)+ x \cdot 2x \cos(x^2) \right ]\)

\(\displaystyle g''(x) = - 2\left [ \sin(x^2)+ 2x^2 \cos(x^2) \right ]\)

\(\displaystyle g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]\)

Example Question #2124 : High School Math

Define \(\displaystyle f(x) = \frac{1}{e^{4x}}\).

What is \(\displaystyle f ''(x)\)?

Possible Answers:

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

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

\(\displaystyle f''(x) =- \frac{16 }{e^{4x}}\)

\(\displaystyle f(x) = \frac{1}{16e^{4x}}\)

\(\displaystyle f''(x) = \frac{16 }{e^{4x}}\)

Correct answer:

\(\displaystyle f''(x) = \frac{16 }{e^{4x}}\)

Explanation:

Rewrite:

\(\displaystyle f(x) = \frac{1}{e^{4x}}\)

\(\displaystyle f(x) = e^{-4x}\)

Take the derivative \(\displaystyle f'\) of \(\displaystyle f\), then take the derivative of \(\displaystyle f'\).

\(\displaystyle f'(x) = -4 e^{-4x}\)

\(\displaystyle f''(x) = -4 \cdot \left (-4 \right )e^{-4x} = 16e^{-4x} = \frac{16 }{e^{4x}}\)

Example Question #13 : Finding Second Derivative Of A Function

What is the second derivative of \(\displaystyle x^2+5x\)?

Possible Answers:

\(\displaystyle \frac{1}{3}x^3+\frac{5}{2}x^2+c\)

\(\displaystyle 10\)

\(\displaystyle 2x+5\)

\(\displaystyle 3x^2+5\)

\(\displaystyle 2\)

Correct answer:

\(\displaystyle 2\)

Explanation:

To get the second derivative, first we need to find the first derivative.

To do that, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=(2*x^{2-1})+(1*5x^{1-0})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5x^0\)

Remember that anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5*1\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5\)

Now we do the same process again, but using \(\displaystyle 2x+5\) as our expression:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(1*2x^{1-1})+(0*5x^{0-1})\)

Notice that \(\displaystyle (0*5x^{0-1})=0\), as anything times zero will be zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(1*2x^{1-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(2x^{0})\)

Anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(2*1)\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=2\)

Example Question #301 : Computation Of The Derivative

What is the second derivative of \(\displaystyle 5x+8\)?

Possible Answers:

Undefined

\(\displaystyle \frac{5}{x}\)

\(\displaystyle 0\)

\(\displaystyle 5\)

\(\displaystyle \frac{1}{5x}\)

Correct answer:

\(\displaystyle 0\)

Explanation:

To get the second derivative, first we need to find the first derivative.

To do that, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat \(\displaystyle 8\) as \(\displaystyle 8x^0\), as anything to the zero power is one.

That means this problem will look like this:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5*x^{1-1})+(0*8x^{0-1})\)

Notice that \(\displaystyle (0*8x^{0-1})=0\) as anything times zero will be zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5*x^{1-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5*x^{0})\)

Remember, anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=5*1\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=5\)

Now to get the second derivative we repeat those steps, but instead of using \(\displaystyle 5x+8\), we use \(\displaystyle 5x^0\).

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^0)=(0*5x^{0-1})\)

Notice that \(\displaystyle (0*5x^{0-1})=0\) as anything times zero will be zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^0)=0\)

Example Question #43 : Derivatives

What is the second derivative of \(\displaystyle x^3+2x+5\)?

Possible Answers:

\(\displaystyle 12x+5\)

\(\displaystyle 18x^4+13x^2+c\)

\(\displaystyle 6x\)

\(\displaystyle 3x^2+2\)

\(\displaystyle 6\)

Correct answer:

\(\displaystyle 6x\)

Explanation:

To get the second derivative, first we need to find the first derivative.

To do that, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat \(\displaystyle 5\) as \(\displaystyle 5x^0\), as anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{3-1})+(1*2x^{1-1})+(0*5x^{0-1})\)

Notice that \(\displaystyle (0*5x^{0-1})=0\), as anything times zero is zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{3-1})+(1*2x^{1-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{2})+(2x^{0})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=3x^2+2\)

Now we repeat the process using \(\displaystyle 3x^2+2\) as the expression.

Just like before, we're going to treat \(\displaystyle 2\) as \(\displaystyle 2x^0\).

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(3x^2+2)=(2*3x^{2-1})+(0*2x^{0-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}3x^2+2=(2*3x^{2-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(3x^2+2)=6x^{1}\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(3x^2+2)=6x\)

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