# Calculus Part 2

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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?
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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.

• Calculus Part 2 -

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.

• Calculus Part 2 -

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