Computation of the Derivative - AP Calculus AB

Card 0 of 3377

Question

Find the derivative.

$\sin (x)$

Answer

The derivative of $\sin (x)$ is $\cos (x)$ . (Memorization)

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Question

Let .

Find the second derivative of .

Answer

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

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

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Question

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

What is ?

Answer

Take the derivative of , then take the derivative of .

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

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

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

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

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Question

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

What is ?

Answer

Take the derivative of , then take the derivative of .

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

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

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

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

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

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Question

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

What is ?

Answer

Rewrite:

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

$f(x) = e^{-4x}$

Take the derivative of , then take the derivative of .

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

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

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Question

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

What is ?

Answer

Take the derivative of , then take the derivative of .

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

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

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

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

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

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

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Question

What is the second derivative of ?

Answer

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.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=(2x^{2-1})+(15x^{1-0})$

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

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+51$

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

Now we do the same process again, but using as our expression:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(12x^{1-1})+(05x^{0-1})$

Notice that $(05x^{0-1})=0$ , as anything times zero will be zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(12x^{1-1})$

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

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(21)$

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

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Question

What is the second derivative of ?

Answer

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 as , as anything to the zero power is one.

That means this problem will look like this:

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{1-1})+(08x^{0-1})$

Notice that $(08x^{0-1})=0$ as anything times zero will be zero.

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

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

Remember, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=51$

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

Now to get the second derivative we repeat those steps, but instead of using , we use .

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^0)=(05x^{0-1})$

Notice that $(05x^{0-1})=0$ as anything times zero will be zero.

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

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Question

What is the second derivative of ?

Answer

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 as , as anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3x^{3-1})+(12x^{1-1})+(05x^{0-1})$

Notice that $(05x^{0-1})=0$ , as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3x^{3-1})+(12x^{1-1})$

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

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

Now we repeat the process using as the expression.

Just like before, we're going to treat as .

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+2)=(23x^{2-1})+(02x^{0-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}3x^2+2=(23x^{2-1})$

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

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

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Question

If , what is ?

Answer

The question is asking us for the second derivative of the equation. First, we need to find the first derivative.

To do that, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})+(08x^{0-1})$

Notice that $08x^{0-1}=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=12x$

Now we do the exact same process but using as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=112x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=12x^{0}$

As stated earlier, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=12*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=12$

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Question

What is the second derivative of ?

Answer

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

To do that, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1x^{1-1})+(0x^{0-1})$

Notice that $(0x^{0-1})=0$ since anything times zero is zero.

That leaves us with $\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1x^{1-1})$ .

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(x^{0})$

As stated earlier, anything to the zero power is one, leaving us with:

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=1$

Now we can repeat the process using or as our equation.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^0)=0x^{0-1}$

As pointed out before, anything times zero is zero, meaning that $\frac{\mathrm{d} }{\mathrm{d} x}(x^0)=0$ .

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Question

What is the second derivative of ?

Answer

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

To do that, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})+(04x^{0-1})$

Notice that $(04x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{1})+(113x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x^{1})+(13x^{0})$

Just like it was mentioned earlier, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x)+(131)$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=24x+13$

Now we repeat the process using as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(124x^{1-1})+(013x^{0-1})$

Like before, anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+10)=(124x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(24x^{0})$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(24*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=24$

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Question

Find the derivative.

$\sin (2x)$

Answer

Use the chain rule to find the derivative: $\frac{d}{dx}f(g(x))=g'(x)f'(g(x))$

Thus, $\frac{d}{dx}\sin (2x)=2\cos (2x)$ .

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Question

Find the derivative of:

$h(t)=\ln(\ln t)$

Answer

On this problem we have to use chain rule, which is:

$\frac{d}{dt}g(f(t))=g'(f(t))f'(t)$

So in this problem we let

$g(t)=\ln(t)$

and

$f(t)=\ln(t)$ .

Since

$g'(t)=\frac{1}{t}$

and

$f'(t)=\frac{1}{t}$ ,

we can conclude that

$h'(t)=\frac{d}{dt}g(f(t))=\frac{1}{f(t)}\cdot\frac{1}{t}=\frac{1}{t\ln t}$

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Question

Determine the derivative of $f(x)=2\tan ^2(x^2)$

Answer

This is a pure problem on understanding how chain rules work for derivatives.

First thing we need to remember is that the derivative of $\tan(x)$ is $\sec^2(x)$ .

When we are taking the derivative of $f(x)=2\tan ^2(x^2)$ , we can first pull out the 2 in the front and we treat $\tan^2(x^2)$ as $[\tan(x^2)]^2$ .

This way, the derivative will become $22\tan(x^2)\frac{\mathrm{d} tan(x^2)}{\mathrm{d} x}$ ,

which is $4\tan(x^2)(2x\sec(x^2))$ .

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Question

$f(x)=sin(x^{2}})$ . Find .

Answer

To take the derivative, you must first take the derivative of the outside function, which is sine. However, the , or the angle of the function, remains the same until we take its derivative later. The derivative of sinx is cosx, so you the first part of will be . Next, take the derivative of the inside function, . Its derivative is , so by the chain rule, we multiply the derivatives of the inside and outside functions together to get .

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Question

. Using the chain rule for derivatives, find .

Answer

By the chain rule, we must first take the derivative of the outside function by bringing the power down front and reducing the power by one. When we do this, we do not change the function that is in the parentheses, or the inside function. That means that the first part of will be . Next, we must take the derivative of the inside function. Its derivative is . The chain rule says we must multiply the derivative of the outside function by the derivative of the inside function, so the final answer is .

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Question

$f(x)=6^(^5^x^{^{2}-7x}^)$ . Find the derivative.

Answer

When the function is a constant to the power of a function of x, the first step in chain rule is to rewrite f(x). So, the first factor of f(x) will be $(6^{5x^2-7x})$ . Next, we have to take the derivative of the function that is the exponent, or . Its derivative is 10x-7, so that is the next factor of our derivative. Last, when a constant is the base of an exponential function, we must always take the natural log of that number in our derivative. So, our final factor will be . Thus, the derivative of the entire function will be all these factors multiplied together: $f'(x)=6^{5x^2-7x}(10x-7)ln6}$ .

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Question

Find the derivative of the function: $y=e^{sinx^3}}$ .

Answer

Whenever we have an exponential function with , the first term of our derivative will be that term repeated, without changing anything. So, the first factor of the derivative will be $e^{sinx^3}$ . Next, we use chain rule to take the derivative of the exponent. Its derivative is . So, the final answer is $y'=e^{sinx^3}cosx^33x^2$ .

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Question

Find the derivative of the exponential function, $y=e^{sec3\theta }$ .

Answer

To take the derivative of any exponential function, we repeat the exponential function in the derivative. So, the first factor of the derivative will be $e^{sec3\theta}$ . Next, we have to take the derivative of the exponent using chain rule. The derivative of the trigonometric function secx is secxtanx, so in terms of this problem its derivative is $sec(3\theta)tan(3\theta)$ . Since the angle has a scalar of 3, we must also multiply the entire derivative by 3. So, the answer is $y'=3e^{sec3\theta}*sec(3\theta)tan(3\theta)$ .

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