Question

1. find the gradient to the normal to the curve x^4-4xy+2y^2=0 at point (-1,3)..
2. differentiate f(x)= (x^3-9)^10 with respect of x

Answer 1

1)We have $F(x,y)=x^4-4xy+2y^2=0$.

The implicit function theorem tells us that gradient of the tangent is $y'(x)=-\frac{F'_x(x,y)}{F'_y(x,y)}$. The tangent and the normal are perpendicular, so gradient of the normal is $-\frac{1}{y'(x)}=\frac{F'_y(x,y)}{F'_x(x,y)}$

$F'_x(x,y)=4x^3-4y$, $F'_y(x,y)=-4x+4y$ , so $\frac{1}{y'(x)}=\frac{y-x}{x^3-y}$ and $-\frac{1}{y'(-1)}=\frac{3-(-1)}{(-1)^3-3}=-1$

Answer: -1

2)$((x^3-9)^{10})'=10(x^3-9)^9\cdot 3x^2=30x^2(x^3-9)^9$