Question

How do you differentiate `y =-2e^(xcosx)` using the chain rule?

Answer 1

The answer is -2`e^(xcosx)` ∗ `cosx-xsinx`.

Explanation:

The `d/dx` of `e^x` is always itself.

Now that we know this we move forward the trick here is that we take the `d/dx` of `e^x` then we apply the chain rule to `e^(xcosx)` we take the `d/dx` of the inside `xcosx`. Here we apply the product rule `f^1(g)` x `g^1(f)` the `d/dx` of this is `cosx-xsinx`. Our final answer is -2`e^(xcosx)` ∗ `cosx-xsinx`. We don't take the 2 into account because its a constant.