Discussion Forum

The  slope of the tangent to the curve   $(y-x^{5})^{2}=x(1+x^{2})^{2}$ at the point (1,3) is 

Answer:

Explanation:

 Plan

 Slope  of tangent  at the point   $(x_{1},y_{1})$  is     $\left(\frac{dy}{dx}\right)_{(x_{1},y_{1})}$

 Given curve                    $(y-x^{5})^{2}=x(1+x^{2})^{2}$

 $\Rightarrow   2(y-x^{5})\left(\frac{\text{d}y}{\text{d}x}-5x^{4}\right)=(1+x^{2})^{2}+2x(1+x^{2}).2x$

  Put x=1, y=3   

$\therefore$           $\frac{dy}{dx}=8$