Question

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

Answer 1

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