Calculus Part 2
posted by Samantha on .
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)/(a0)
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.

y = tan u
dy/dx = sec^2 u du/ dx
here u = x^(1/2)
du/dx = sec^2 (x^.5) .5 x^.5
=(1/2) (1/sqrt x) sec^2(sqrt x)
I agree with the authorities. 
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)/(a0)
I'm not sure how to solve for a.

sketching that I see no such line. Are you sure it is not through (1/4 , 0)