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To find \(\displaystyle f'(x)\) first find \(\displaystyle f(x)=g(h(x))\).

Since,

\(\displaystyle g(x)=x^{3}+x^{2}-1\)

\(\displaystyle h(x)=(x^{2}-2)\)

then,

\(\displaystyle f(x)=(x^2-2)^3+(x^2-2)^2-1\).

We have to use the chain rule and power rule to solve this problem:

Chain Rule: \(\displaystyle f'(x)=g'(h(x))\cdot(h'(x))\)

Power Rule: \(\displaystyle (x^n)'=nx^{n-1}\)

Therefore,

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

\(\displaystyle f'(x)=6x(x^2-2)^2+4x(x^2-2)\)

To solve this problem, we must first find f'(t) by deriving the original equation.

In this case, we just need to use the power rule which states,

\(\displaystyle [f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)\). 

\(\displaystyle {f}(t)= 2t^{3}-3t\)

\(\displaystyle f{}'(t)= 6t^{2}-3\)

We can then derive f'(t), again using just the power rule which states \(\displaystyle (x^n)'=nx^{n-1}\).

\(\displaystyle f''(t)= 12t\)

When we plug 4 into this final function, we arrive at the answer:

\(\displaystyle f''(4)= 12(4)=48\)

Taking the first derivative gives us the function for the slope for our entire equation. Thus, finding where the slope is positive tells us where the function is increasing.

To find the general solution for the separable differential equation, we must move x and dx, y and dy to the same sides:

\(\displaystyle \frac{dy}{y}=(x^3+x^2)dx\)

Next, we integrate both sides:

\(\displaystyle \ln\left | y\right |=\frac{x^3}{3}+\frac{x^4}{4}+C\)

The integrals were found using the following rules:

\(\displaystyle \int \frac{dx}{x}=\ln\left | x\right |+C\), \(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\)

To solve for y, we must exponentiate both sides of the equation:

\(\displaystyle y=e^{\frac{x^3}{3}+\frac{x^4}{4}+C}=e^{\frac{x^3}{3}+\frac{x^4}{4}}\cdot e^C=Ce^{\frac{x^3}{3}+\frac{x^4}{4}}\)

To solve the separable differential equation, we must move x and dx, y and dy to their respective sides, and integrate both sides:

\(\displaystyle e^y dy=x^3 dx\)

\(\displaystyle \int e^y dy=e^y+C\)

\(\displaystyle \int x^3 dx= \frac{x^4}{4}+C\)

The integrations were performed using the following rules:

\(\displaystyle \int e^x dx=e^x+C\)

\(\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C\)

The equation then becomes

\(\displaystyle y=\frac{x^4}{4}+C\)

Note that we combine all constants of integration to a single C.

Now, we solve for y by taking the natural logarithm of both sides of the equation:

\(\displaystyle y=\ln(\frac{x^4}{4}+C)\)

Simplifying, we get

\(\displaystyle y=C\ln(\frac{x^4}{4})\).

Using integrating factors:

\(\displaystyle e^{(-t)}*(y'-y)=0\) 

Once we use this integrating factor, we can reduce the equation to:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}[e^{-t}y]=0\)

Integrating both sides: 

\(\displaystyle e^{-t}y=C\)

\(\displaystyle y=Ce^{t}\)

Since you may or may not know how to solve \(\displaystyle 2nd\) order Differential equations, you can always plug in the answers and see if it is correct. 

Let's start with \(\displaystyle y=cos(t)\): 

\(\displaystyle y'=-sin(t)\)

\(\displaystyle y''=-cos(t)\)

\(\displaystyle y''-y=-cos(t)-cos(t)=-2cos(t)\neq 0\)

If you try all the other function, they will solve the differential equation. 

This involves solving a differential equation for \(\displaystyle V\). 

Firstly, we determine the value of \(\displaystyle \frac{d}{dt}Q\):

\(\displaystyle Q=t^2-t+2\)

\(\displaystyle \frac{d}{dt}Q=2t-1\)

At \(\displaystyle t=5\)

\(\displaystyle \frac{d}{dt}Q(5)=2*5-1=9\)

This means our equation is now:

\(\displaystyle \frac{d}{dx}V =\frac{d}{dt}Q*R\)

\(\displaystyle \frac{d}{dx}V =9*5=45\)

To determine the difference of \(\displaystyle V\) from \(\displaystyle x:(0,1)\), we take the definite integral:

\(\displaystyle V=\int_{0}^{1}45 dx= 45-0=45\)

The derivative of the trigonometric function \(\displaystyle \sin(x)\) is \(\displaystyle \cos(x)\).

Therefore, the correct answer is 

\(\displaystyle g'(x)= 5\cos(x)\).