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 & explanation
Correct answer: option 3
The equation of the curve is
$\left(y-x^5\right)^2=x\left(1+x^2\right)^2$
Differentiating both sides with respect to x, we get
$2\left(y-x^5\right)\left(\frac{d y}{d x}-5 x^4\right)=\left(1+x^2\right)^2+4 x^2\left(1+x^2\right)$
Putting x = 1, y = 3 on both sides, we get
$4\left(\frac{d y}{d x}-5\right)=4+8 \Rightarrow \frac{d y}{d x}=8$