Application of Derivatives 1 Question 22

22. The slope of the tangent to the curve (yx5)2=x(1+x2)2 at the point (1,3) is

(2014 Adv.)

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Solution:

  1. Slope of tangent at the point (x1,y1) is dydx(x1,y1).

Given curve, (yx5)2=x(1+x2)2

2(yx5)dydx5x4=(1+x2)2+2x(1+x2)2x

Put x=1 and y=3,dy/dx=8



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