March 29, 2015

Homework Help: Calculus Part 2

Posted by Samantha on Tuesday, November 4, 2008 at 8:01pm.

y = tan(sqrtx)
Find dy/dx.
So I found it and I got the answer as sec^2(sqrtx) but the answer is (sec^2(sqrtx))/2(sqrtx)...why?

2) Find the line which passes through the point (0, 1/4) and is tangent to the curve y=x^3 at some point.
So I found the derivative which is 3x^2.
Let (a, a3) be the point of tangency.

3x^2 = (a3 - 1/4)/(a-0)
I'm not sure how to solve for a.

Can someone please help me out with these? I have an exam tomorrow evening and these questions are a lost cause.

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