How to Calculate the gradient of $y=(4-2x^2)^5$ at the point where $x=1$

Question

I think that I would have to use chain rule here and I did and I got $dy/dx=5(-4x) (4-2x^2)^4$

but I have no idea how to use the gradient in this case...like do I expand my brackets or do I just leave the equation as is and put it equal to zero...Im so confused please help

Answer 1

Just use the Chain Rule.

$$\frac{dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx}$$

$$y = f(u) = u^5 = (4-2x^2)^5$$

$$u = g(x) = 4-2x^2$$

$$\frac{dy}{dx} = 5(4-2x^2)^4\cdot(-4x)$$

You’ve gotten that right.

No expansion or simplification is required. You are given a value for $x$: $1$.

Just enter $x = 1$ and you’ll get the answer, which becomes $-320$.