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)/(a-0)
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)
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