How do i use the chain rule twice when differentiating?

Ok so for this question we'll use this example: y = (e-2x^2 + 2)1/5
We start off by making the expression inside the brackets equal to u. In other words u = e-2x^2 + 2. This also means y = u1/5
Using the chain rule, we need to find du/dx and dy/du. dy/du = 1/5u-4/5. Now to find du/dx we must use the chain rule again as e-2x^2 has a power within a power.
So lets say z = e-2x^2 . You can use any letter but for this example i'll use z. And v = -2x2. Using the chain rule we get: dz/dx = -4xe-2x^2
Now we can go back to when we used the chain rule the first time to differentiate the whole expression. du/dx = -4xe-2x^2 and dy/du = 1/5u-4/5. Multiplying these together and replacing u with e-2x^2 + 2 gives us our final answer. dy/dx = 1/5(e-2x^2 + 2)-4/5-4xe-2x^2
Simplifying gives us dy/dx = -4xe-2x^2/5(e-2x^2 + 2)4/5

Answered by Indiya L. Maths tutor

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