Question
Question: How do you differentiate the given function \(\arcsin \left( 2x \right)\)?...
How do you differentiate the given function arcsin(2x)?
Solution
We start solving the problem by assuming 2x=z and then differentiating both sides of the given function with respect to x. We then recall the chain rule of differentiation as dxd(g(f))=dfd(g)×dxdf to proceed through the problem. We then make use of the fact that dxd(arcsin(x))=1−x21 to proceed through the problem. We then make use of the facts that dxd(ax)=a to get the required answer for the derivative of the function.
Complete step-by-step answer:
According to the problem, we are asked to find the derivative of the function arcsin(2x).
Let us assume y=arcsin(2x) ---(1).
Let us assume 2x=z. Let us substitute this in equation (1).
⇒y=arcsin(z) ---(2).
Let us differentiate both sides of the equation (2) with respect to x.
⇒dxdy=dxd(arcsin(z)) ---(3).
From chain rule of differentiation, we know that dxd(g(f))=dfd(g)×dxdf. Let us substitute this result in equation (3).
⇒dxdy=dzd(arcsin(z))×dxdz ---(4).
We know that dxd(arcsin(x))=1−x21. Let us use this result in equation (4).
⇒dxdy=1−z21×dxdz ---(5).
Now, let us substitute z=2x in equation (5).
⇒dxdy=1−4x21×dxd(2x) ---(6).
We know that dxd(ax)=a. Let us use this result in equation (6).
⇒dxdy=1−4x22.
∴ We have found the derivative of the function arcsin(2x) as 1−4x22.
Note: Whenever we get this type of problem, we try to make use of chain rule to get a solution to the given problem. We should not forget to different 2x after performing equation (5) which is the common mistake done by students. We can also solve this problem by making use of the fact that dxd(arcsin(f(x)))=1−(f(x))2dxd(f(x)) to get the required answer. Similarly, we can expect problems to find the derivative of the function y=arctan(log(5x)).