The slope of the tangent to the curve $\left(y-x^{5}\right)^{2}=x\left(1+x^{2}\right)^{2}$ at the po

Question

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

Professor PC Thomas Classes · Answer

(8) : $\left(y-x^{5}ight)^{2}=x\left(1+x^{2}ight)^{2}$ Differentiating with respect to $x$, we get $2\left(y-x^{5}ight)\left(y_{1}-5 x^{4}ight)=\left(1+x^{2}ight)^{2}+x \cdot 2\left(1+x^{2}ight) \cdot 2 x$ At the point $(1,3)$\ $2(3-1)\left(y_{1}-5ight)=4+8=12$\ $	herefore y_{1}=\frac{12}{4}+5=8$\