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:

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

We can first recognize that 

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ 

since $\displaystyle dt$ cancels out when we divide.

Then,  given $\displaystyle x=2t+13$ and $\displaystyle y=4t+17$ and using the Power Rule

$\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$ for all $\displaystyle n\neq0$,

we can determine that 

$\displaystyle \frac{dy}{dt}=(1)4t^{1-1}+(0)17=4$ and $\displaystyle \frac{dx}{dt}=(1)2t^{1-1}+(0)13=2$.

Therefore, 

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{4}{2}=2$.

We can first recognize that 

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ 

since $\displaystyle dt$ cancels out when we divide.

Then,  given $\displaystyle x=-t+12$ and $\displaystyle y=\frac{8}{t}$ or $\displaystyle y=8t^{-1}$ and using the Power Rule

$\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$ for all $\displaystyle n\neq0$,

we can determine that 

$\displaystyle \frac{dy}{dt}=(-1)8t^{-1-1}=-\frac{8}{t^{2}}$ and $\displaystyle \frac{dx}{dt}=(1)(-1)t^{1-1}+(0)12=-1$.

Therefore, 

$\displaystyle \frac{dy}{dx}=\frac{-\frac{8}{t^{2}}}{-1}=\frac{8}{t^{2}}$.

We can first recognize that

 $\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ 

since $\displaystyle dt$ cancels out when we divide.

Then,  given $\displaystyle x=-6t+10$ and $\displaystyle y=7t-2$ and using the Power Rule

$\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$ for all $\displaystyle n\neq0$,

we can determine that 

$\displaystyle \frac{dy}{dt}=(1)7t^{1-1}-(0)2=7$ and $\displaystyle \frac{dx}{dt}=(1)(-6)t^{1-1}+(0)10=-6$.

Therefore, 

$\displaystyle \frac{dy}{dx}=\frac{7}{-6}=-\frac{7}{6}$..

The first derivative of a parametrically defined curve is found by computing

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

We need to find the derivatives of y(t) and x(t) separately, and then find the quotient of the derivatives.

You will need to know that 

$\displaystyle \frac{d}{dt}sin(t)=cos(t)$ and that $\displaystyle \frac{d}{dt}t^n=nt^{n-1}$.

$\displaystyle x'(t)=\frac{dx}{dt}=3cos(t)$

$\displaystyle y'(t)=\frac{dy}{dt}=2t+4$

Thus,

 $\displaystyle \frac{dy}{dx}=\frac{2t+4}{3cos(t)}$

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

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

The derivative of the equation for $\displaystyle y$ is

$\displaystyle \frac{dy}{dt}=2$

and the derivative of the equation for $\displaystyle x$ is

$\displaystyle \frac{dx}{dt}=3t^2+4t+4$

The derivatives were found using the following rule:

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

The derivative of a parametric function is given by

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

So, we must find the derivative of the functions with respect to t:

$\displaystyle y'=e^t+te^t-2\cos(t)\sin(t)$, $\displaystyle x'=3$

The derivatives were found using the following rules:

$\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}$, $\displaystyle \frac{d}{dx}(e^x)=e^x$ $\displaystyle \frac{d}{dx}\cos(t)=-\sin(t)$, $\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

Simply divide the derivatives to get your answer. 

We can first recognize that

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ 

since $\displaystyle dt$  cancels out when we divide.

Then, given  $\displaystyle x=10t-3$ and $\displaystyle y=2t+9$ and using the Power Rule

$\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$ for all $\displaystyle n\neq0$, we can determine that 

$\displaystyle \frac{dx}{dt}=(1)10t^{1-1}-(0)3=10$ and 

$\displaystyle \frac{dy}{dt}=(1)2t^{1-1}+(0)9=2$.

Therefore, 

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{10}=\frac{1}{5}$.

We can first recognize that 

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ 

since $\displaystyle dt$  cancels out when we divide.

Then, given $\displaystyle x=3t+2$ and $\displaystyle y=9t-5$ and using the Power Rule

$\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$ for all $\displaystyle n\neq0$, we can determine that 

$\displaystyle \frac{dx}{dt}=(1)3t^{1-1}+(0)2=3$ and 

$\displaystyle \frac{dy}{dt}=(1)9t^{1-1}-(0)5=9$.

Therefore,

 $\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{9}{3}=3$.

We can first recognize that

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ 

since $\displaystyle dt$  cancels out when we divide.

Then, given $\displaystyle x=5t-1$ and $\displaystyle y=t+5$ and using the Power Rule

$\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$ for all $\displaystyle n\neq0$, we can determine that

$\displaystyle \frac{dx}{dt}=(1)5t^{1-1}-(0)1=5$  and

$\displaystyle \frac{dy}{dt}=(1)t^{1-1}+(0)5=1$ .

Therefore, 

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{1}{5}$.