Question

find the slope of the curve 4x^2y+16 at the point (2,1)

Answer 1

Find its derivative,

4[2x^2y(y/x + log(x)dy/dx)]
(8yx^2y/x) + (8x^2y.log(x) dy/dx) = dy/dx
(8yx^2y/x) = dy/dx(1 - 8x^2y.log(x) )

dy/dx = (8yx^2y/x) /(1 - 8x^2y.log(x))

So putting the values of x and y (2,1)

dy/dx = [(8)(1)(2^2)/2]/[1 - (8)(2^2)(log(2))]
= 16 / (1-32 log 2)
=16/-21.1
=-0.755