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 ______

Answer 1

Answer: 8

$2\left(y-x^5\right)\left(\frac{dy}{dx}-5x^4\right)$
$=1{\left(1+x^2\right)}^2+(x)\left(2\left(1+x^2\right)(2x)\right)$
Now put $x=1,y=3$ and $\frac{dy}{dx}=m$.
$2(3-1)(m-5)=1(4)+(1)(4)(2)$
$m-5=\frac{12}{4}$
$m=5+3=8$
$\frac{dy}{dx}=m=8$