# Math Help - [SOLVED] Find an equation of the tangent to the curve

1. ## [SOLVED] Find an equation of the tangent to the curve

Directions: Find an equation of the tangent to the curve at the point corresponding to the value of the parameter.

x=e^√t , y= t-ln t^9 ; t = 1

Also, please be sure to add captions under each step so that I am able to understand each part of the process. Thankyou

2. Hint: The relevant rules of differentiation here are:

\begin{aligned}
\frac{d}{dx}x^n &= nx^{n-1}\\
\frac{d}{dx}e^x &= e^x.
\end{aligned}

The Chain Rule states,

$\frac{d}{dx}f(g(x))=f'(g(x))g'(x).$

The slope of the tangent line can be found with the formula

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt},$

and the formula for a line passing through $(x_1,y_1)$ with slope $m$ is

$y-y_1=m(x-x_1).$

Hope this helps!

3. Originally Posted by Inexcess
Directions: Find an equation of the tangent to the curve at the point corresponding to the value of the parameter.

x=e^√t , y= t-ln t^9 ; t = 1

Also, please be sure to add captions under each step so that I am able to understand each part of the process. Thankyou
If you are working on problems involving parametric equations, you really should already know the basic derivatives Scott H gave, but it really is that basic!

The tangent line to a curve corresponding to y= f(x), at $(x_0,y_0)$m is given by $y= m(x- x_0)+ f(x_0)$ where $m= \frac{dy}{dx}(x_0)$.

Here, you cannot find $\frac{dy}{dx}$ directly, but you can use the fact that $\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, which follows from the chain rule.

4. Hey, guys. I finally figured it out, haha. It's pretty bad when someone is so stressed out that they can't remember something as simple as the chain rule. That's all that was messing me up, haha. All I had to do was take a break from work, and I came right back to it and figured it out on the spot. I appreciate it even though I feel pretty dumb, now. Damn you, school! Stress is not helpful!