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

2 Answers
kumail
May 10, 2017

#-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)#

Shwetank Mauria
May 10, 2017

#(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#

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