Question

Differentiate `1/sinx^2` using chain rule?

Answer 1

`-2xcos(x^2)/sin^2(x^2)`

Explanation:

We can use the chain rule to differentiate the function, the chain rule states `d/dx f(g(x)) = f'(g(x)) * g'(x)`

So `d/dx (sin(x^2))^-1 = -1sin^-2(x^2)*cos(x^2)*2x = -2xcos(x^2)/sin^2(x^2)`

Answer 2

`(df)/(dx)=-2xcotx^2csc^2x^2`

Explanation:

We use chain rule here.

Using this in order to differentiate a function of a function, say `y, =f(g(x))`, where we have to find `(dy)/(dx)`, we need to do (a) substitute `u=g(x)`, which gives us `y=f(u)`. Then we need to use a formula called Chain Rule, which states that `(dy)/(dx)=(dy)/(du)xx(du)/(dx)`.

In fact if we have something like `y=f(g(h(x)))`, we can have `(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)`

Here we have `f(x)=1/(g(x))`, where `g(x)=sin(h(x))` and `h(x)=x^2`.

Hence, `(df)/(dx)=-1/(sin^2x^2)xxcosx^2xx2x=-2xcotx^2csc^2x^2`