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, 

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