3.) Give the equation of the line tangent to the curve at the given point.(adsbygoogle = window.adsbygoogle || []).push({});

a.) (y)(tan^-1 x) = x*y at (sqrt(3),0)

b.) ln y = x^2 +(2)*e^x at (0, e^2)

Please help me with this problem, I am stuck.

Here is some of my work process:

i have no idea how to solve for y in a.)

but for b.) I took the expontential of both sides

then i got y= e^(x^2) + e^(2*e^x)

do I find the derivate of y to get the slope

but how do i plug in the values of the pt.

do i use the y- yl= m(x-x1) formula to find the tangent to the curve?

please help me with these problems, I'm clueless for part a, and i have a bit of an understanding on how to do part b.)

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# Equation of Tangent to the Line

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