We start by taking the derivative of x and y with respect to t, as both of the equations are only in terms of this variable:

\(\displaystyle \frac{dx}{dt}=6t^2+10t-11\)

\(\displaystyle \frac{dy}{dt}=14t-4\)

The problem asks us to find the derivative of the parametric equations, dy/dx, and we can see from the work below that the dt term is cancelled when we divide dy/dt by dx/dt, leaving us with dy/dx:

\(\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dx}\)

So now that we know dx/dt and dy/dt, all we must do to find the derivative of our parametric equations is divide dy/dt by dx/dt:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{14t-4}{6t^2+10t-11}=\frac{2(7t-2)}{6t^2+10t-11}\)

The integration involves breaking up a power of a trigonometric ratio, and then using known trigonometric identities.

\(\displaystyle \int sin^3x cos^4x dx = \int [sin x \cdot sin^2x \cdot cos^4x]dx\)\(\displaystyle = \int \left[sin x \cdot \left(\frac{1}{2} - \frac{1}{2}cos 2x\right)\cdot cos^4x\right] dx\)

\(\displaystyle = \frac{1}{2}\int [sin x cos^4x - sin x cos 2x cos^4x]dx\)

\(\displaystyle = \frac{1}{2} \int [sin x cos^4x - sin x (2cos^2x - 1)cos^4x ]dx\)

\(\displaystyle = \frac{1}{2} \int [2sin x cos^4x - 2sin x cos^6x]dx\)

\(\displaystyle = \frac{1}{2} \cdot -\frac{1}{5} \cdot 2cos^5x - \frac{1}{2} \cdot 2 \cdot- \frac{1}{7}cos^7x + C\)

\(\displaystyle = -\frac{1}{5} cos^5x + \frac{1}{7} cos^7x + C\)

The alternative is to find which answer choice has a derivative equal to the answer choice, and for this we get:

\(\displaystyle {}\frac{\mathrm{d} }{\mathrm{d} x}\left(-\frac{1}{5} cos^5x + \frac{1}{7} cos^7x + C\right) = cos^4x\cdot sin(x) - cos^6x\cdot sin(x)\)

\(\displaystyle {}= cos^4xsin(x)[1 - cos^2x] = cos^4x\cdot sin(x)\cdot sin^2x = sin^3xcos^4x\)

Write the the formula to solve for the derivative of parametric functions.

\(\displaystyle \large \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

Find \(\displaystyle \frac{dy}{dt}\)and \(\displaystyle \frac{dx}{dt}\) using the power rule \(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\).

\(\displaystyle \frac{dy}{dt}=12t^3\)

\(\displaystyle \frac{dx}{dt}=6t^2\)

Substitute back to the formula to obtain the derivative.

\(\displaystyle {} \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{12t^3}{6t^2}=2t\)

The derivative of a parametric function is given by:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

where

\(\displaystyle \frac{dy}{dt}=\cos(t)-\sin(t)\), \(\displaystyle \frac{dx}{dt}=2t\)

The derivatives were found using the following rules:

\(\displaystyle \frac{d}{dx}\cos(x)=-\sin(x)\)

\(\displaystyle \frac{d}{dx}\sin(x)=\cos(x)\)

\(\displaystyle \frac{d}{dx}(x^n)nx^{n-1}\)

Simply divide the derivatives as shown above.

Given equations for \(\displaystyle y\) and \(\displaystyle x\) in terms of \(\displaystyle t\), we can find the derivative of parametric equations as follows:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), as the \(\displaystyle dt\) terms will cancel out.

Using the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) and given \(\displaystyle y=3t^{2}-2\) and \(\displaystyle x=4t^{2}+6\),

\(\displaystyle \frac{dy}{dt}=(2)3t^{2-1}-(0)2=6t\)  and \(\displaystyle \frac{dx}{dt}=(2)4t^{2-1}+(0)6=8t\).

Therefore, 

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{6t}{8t}=\frac{3}{4}\).

The derivative of a parametric equation is given by the following equation:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

Solving for the derivative of the equation for y, you get

\(\displaystyle \frac{dy}{dt}=2t\)

and for the equation for x, you get

\(\displaystyle \frac{dx}{dt}=2-\sin(t)\)

The following rules were used for the derivatives:

\(\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}\), \(\displaystyle \frac{d}{dx}\cos(x)=-\sin(x)\)

Write the formula to find the derivative for parametric equations.

\(\displaystyle \frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}\)

\(\displaystyle \frac{dx}{dt}= -sin(t)\)

\(\displaystyle \frac{dy}{dt}= cos(t)\)

Substitute the knowns into the formula.

\(\displaystyle \frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{-sin(t)}{cos(t)}=-tan(t)\)

We can determine that \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)  since the \(\displaystyle dt\) terms will cancel out in the division process.

Since \(\displaystyle x=5t+9\) and \(\displaystyle y=2t+7\),  we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to derive

\(\displaystyle \frac{dx}{dt}=(1)5x^{1-1}=5\)and \(\displaystyle \frac{dy}{dt}=(1)2t^{1-1}+(0)7=2\).

Thus:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{5}\).

We can determine that \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)  since the \(\displaystyle dt\) terms will cancel out in the division process.

Since \(\displaystyle x=2t-5\) and \(\displaystyle y=3t+8\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to derive

\(\displaystyle \frac{dx}{dt}=(1)2t^{1-1}-(0)5=2\)and \(\displaystyle \frac{dy}{dt}=(1)3t^{1-1}+(0)8=3\) .

Thus:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3}{2}}\).

We can determine that \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)  since the \(\displaystyle dt\) terms will cancel out in the division process.

Since \(\displaystyle x=9t+1\) and \(\displaystyle y=4t+10\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to derive 

\(\displaystyle \frac{dx}{dt}=(1)9t^{1-1}+(0)1=9\) and \(\displaystyle \frac{dy}{dt}=(1)4t^{1-1}+(0)10=4\)  .

Thus:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{4}{9}}\).