Application of Derivatives 1 Question 22
22. 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
(2014 Adv.)
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Solution:
- Slope of tangent at the point $\left(x _1, y _1\right)$ is $\frac{d y}{d x}{ } _{\left(x _1, y _1\right)}$.
Given curve, $\left(y-x^{5}\right)^{2}=x\left(1+x^{2}\right)^{2}$
$\Rightarrow 2\left(y-x^{5}\right) \frac{d y}{d x}-5 x^{4}=\left(1+x^{2}\right)^{2}+2 x\left(1+x^{2}\right) \cdot 2 x$
Put $x=1$ and $y=3, d y / d x=8$
