# Find tangent line at given points, no function equation

I have never encountered a problem like this and am a bit confused.

Function $f$ satisfies:

$f(3)=5$, $f(9)=7$

$f'(3)=11$, $f'(9)=13$

Find an equation for the tangent line to the curve $y=f(x^2)$ at point $(x,y) = (3,7)$.

I don't know where to start. I drew a graph of the points related to the slope of the derivative but that doesn't really help me, and I don't understand the concept of $y = f(x^2)$.

A point in the right direction would be great, thank you for any help!

## 1 Answer

Hint 1: By the Chain Rule, we have $\frac{dy}{dx}=(2x)f'(x^2)$.

Hint 2: The information about $f(3)$ and $f'(3)$ is meant to lead you astray. Not nice!

• OH! So basically it involves implicit differentiation, I derive the function, plug coordinates in as usual. I will try it and come back with result in a second. Thank you! – mrybak834 Oct 7 '14 at 4:47
• It is not really implicit differentiation, though that too usually involves the Chain Rule. It is about finding the slope of the tangent line. The rest will be familiar. Do tell me what your answer is, and I can confirm. – André Nicolas Oct 7 '14 at 4:49
• Tried it really fast, if not right, thank you for the help anyways, I don't want to take up any more of your time and feel like I can figure it out. 2(3)*(f'(9)=6*13=78 therefore y-7=(78)(x-3) – mrybak834 Oct 7 '14 at 4:53
• That's right. You may be expected to simplify. – André Nicolas Oct 7 '14 at 4:54
• Awesome! Thank you very very much! – mrybak834 Oct 7 '14 at 4:55