Question

How do you differentiate `y=x^2tan(1/x)`?

Answer 1

Use a combination of the product rule and the chain rule to get
`dy/dx=2xtan(1/x)-sec^2(1/x)`.

Explanation:

Let `f` and `g` be functions of `x`.
The product rule states that if `y = f*g` then `y'=f'*g+f*g'`.
The chain rule states that if `y=f(g)` then `y'=f'(g)*g'`.

In this question, say `f(x)=x^2`, `g(x)=tan(x)`, and `h(x)=1/x`.

Then `y=f(x)*g[h(x)]`

And so
`y'=[f(x)]'g[h(x)]+f(x){g[h(x)]}'`
`y'=f'(x)*g[h(x)]+f(x)*g'[h(x)]*h'(x)`

This means if `y=x^2tan(1/x)`
then

`dy/dx=d/dx(x^2)*tan(1/x)+x^2*d/dxtan(1/x)`

`dy/dx=2xtan(1/x)+x^2*sec^2(1/x)*d/dx(1/x)`

`dy/dx=2xtan(1/x)+cancel(x^2)sec^2(1/x)*(-1)/cancel(x^2)`

`dy/dx=2xtan(1/x)-sec^2(1/x)`

That's likely as far as you'll need to go.