I tried as hard as I can but couldn't figure out. Help please.
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You can't prove that because it isn't true.
Perhaps what you mean is that you want to prove that d/dx tan^(-1)(x) = 1/(1 + x²).
You can use the chain rule. Suppose that the arctangent function has a derivative. Notice that
tan(tan^(-1)(x)) = x for all real numbers x.
because the tangent and arctangent are inverse functions.
Take the derivative of both sides. On the right, you get d/dx x = 1. On the left, use the chain rule.
[d/dx tan^(-1)(x)] sec²(tan^(-1)(x)) = 1.
Divide out by sec²(tan^(-1)(x)) to get
d/dx tan^(-1)(x) = 1/sec²(tan^(-1)(x)) = 1/[1 + tan²(tan^(-1)(x))] = 1/(1 + x²).
Perhaps what you mean is that you want to prove that d/dx tan^(-1)(x) = 1/(1 + x²).
You can use the chain rule. Suppose that the arctangent function has a derivative. Notice that
tan(tan^(-1)(x)) = x for all real numbers x.
because the tangent and arctangent are inverse functions.
Take the derivative of both sides. On the right, you get d/dx x = 1. On the left, use the chain rule.
[d/dx tan^(-1)(x)] sec²(tan^(-1)(x)) = 1.
Divide out by sec²(tan^(-1)(x)) to get
d/dx tan^(-1)(x) = 1/sec²(tan^(-1)(x)) = 1/[1 + tan²(tan^(-1)(x))] = 1/(1 + x²).