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Question: What is the derivative of \(y = 2{\tan ^{ - 1}}\left( {{e^x}} \right)\) with respect to \(x\)....

What is the derivative of y=2tan⁡−1(ex)y = 2{\tan ^{ - 1}}\left( {{e^x}} \right) with respect to xx.

Explanation

Solution

In the given problem, we are required to differentiate y=2tan⁡−1(ex)y = 2{\tan ^{ - 1}}\left( {{e^x}} \right) with respect to x. Since, y=2tan⁡−1(ex)y = 2{\tan ^{ - 1}}\left( {{e^x}} \right) is a composite function. So, differentiation of y=2tan⁡−1(ex)y = 2{\tan ^{ - 1}}\left( {{e^x}} \right) with respect to x will be done layer by layer using the chain rule of differentiation. Also the derivative of tan⁡−1(x){\tan ^{ - 1}}(x)with respect to xx must be remembered.

Complete step by step answer:
Now, ddx(2tan⁡−1(ex))\dfrac{d}{{dx}}\left( {2{{\tan }^{ - 1}}\left( {{e^x}} \right)} \right). Taking the constant outside the differentiation in order to apply the chain rule of differentiation. So, we get,
2ddx(tan⁡−1(ex))2\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {{e^x}} \right)} \right)
Now, Let us assume u=exu = {e^x}. So substituting ex{e^x} as uu, we get,
2ddx(tan⁡−1(u))2\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( u \right)} \right)
Now ,we know that derivative of inverse tangent function tan⁡−1(x){\tan ^{ - 1}}\left( x \right) with respect to x is (11+x2)\left( {\dfrac{1}{{1 + {x^2}}}} \right). So, we get (21+u2)(dudx)\left( {\dfrac{2}{{1 + {u^2}}}} \right)\left( {\dfrac{{du}}{{dx}}} \right).

Now, putting back uuas ex{e^x}, we get,
(21+(ex)2)(d(ex)dx)\left( {\dfrac{2}{{1 + {{\left( {{e^x}} \right)}^2}}}} \right)\left( {\dfrac{{d\left( {{e^x}} \right)}}{{dx}}} \right) because dudx=d(ex)dx\dfrac{{du}}{{dx}} = \dfrac{{d({e^x})}}{{dx}}
Now, we know that the derivative of exponential function ex{e^x} with respect to x is ex{e^x}. So, we get,
(21+e2x)(ex)\left( {\dfrac{2}{{1 + {e^{2x}}}}} \right)\left( {{e^x}} \right)
Simplifying the expression, we get,
∴2ex1+e2x\therefore \dfrac{{2{e^x}}}{{1 + {e^{2x}}}}

Therefore, the derivative of y=2tan⁡−1(ex)y = 2{\tan ^{ - 1}}\left( {{e^x}} \right) with respect to x is (2ex1+e2x)\left( {\dfrac{{2{e^x}}}{{1 + {e^{2x}}}}} \right).

Note: The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer and hence, the derivative of a composite function can be found using the same rule.