Find the equation of tangent and normal to the curve $y=5x^4$ at the point $(1,5)$

Given $P(1,5)$

Curve $y=5x^4$

Diff w.r.t to ‘$x$’

$\frac{dy}{dx}=5\left(4x^3\right)=20x^3$

slope $\left(m\right)={\left(\frac{dy}{dx}\right)}_{P(1,5)}=20(1{)}^3=20$

i) Equation of tangent

$y-y_1=m\left(x-x_1\right)$

$y-5=20\left(x-1\right)$

$y-5=20x-20$

$20x-y-15=0$

ii) Equation of normal

$y-y_1=\frac{-1}{m}\left(x-x_1\right)$

$y-5=\frac{-1}{20}(x-1)$

$20y-100=-x+1$

$x+20y-101=0$