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March 6, 2015

March 6, 2015

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

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 -
**Damon**, Tuesday, November 4, 2008 at 8:41pmy = 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 -
**Damon**, Tuesday, November 4, 2008 at 8:47pm2) 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)

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