Use $\displaystyle u$-substitution so that $\displaystyle u(x) = 2x+5$.

Then the function $\displaystyle f(x)$ becomes $\displaystyle f(x) = u^2$.

By the chain rule, $\displaystyle f'(x) = \frac{du}{dx} \times \frac{df}{du}$.

We calculate each term using the power rule:

$\displaystyle \frac{du}{dx} = 2$

$\displaystyle \frac{df}{du} = 2u$

Plug in $\displaystyle u=2x+5$:

$\displaystyle f'(x) = 4u = 4 (2x+5) = 8x+20$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $\displaystyle f'$.

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

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

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.

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

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

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.

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

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 $\displaystyle 2x^0$.

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

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

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

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

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 $\displaystyle 8$ as $\displaystyle 8x^0$ since anything to the zero power is one.

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

Notice that $\displaystyle 0*8x^{0-1}=0$ since anything times zero is zero.

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

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

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

Now we do the exact same process but using $\displaystyle 12x$ as our expression.

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

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

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

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

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

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 $\displaystyle 1$ as $\displaystyle 1x^0$ since anything to the zero power is one.

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

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

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

Simplify.

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

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

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

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

Now we can repeat the process using $\displaystyle 1$ or $\displaystyle x^0$ as our equation.

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

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