Solveeit Logo
Login

Question

Question: Differentiate \({\tan ^{ - 1}}(\dfrac{{1 + 2x}}{{1 - 2x}})\)with respect to \(\sqrt {1 + 4{x^2}} .\)...

Differentiate tan⁡−1(1+2x1−2x){\tan ^{ - 1}}(\dfrac{{1 + 2x}}{{1 - 2x}})with respect to 1+4x2.\sqrt {1 + 4{x^2}} .

Explanation

Solution

Hint: In these types of questions assume the first function tan⁡−1(1+2x1−2x){\tan ^{ - 1}}(\dfrac{{1 + 2x}}{{1 - 2x}}) be u and second function 1+4x2\sqrt {1 + 4{x^2}} be v. Now according to question, you need to find dudv\dfrac{{du}}{{dv}}. Nowdudv\dfrac{{du}}{{dv}} can also be written as dudxdvdx\dfrac{{\dfrac{{du}}{{\dfrac{{dx}}{{dv}}}}}}{{dx}}. So, differentiate functions u and v separately and then find the value of dudv\dfrac{{du}}{{dv}} to obtain the solution to the problem.

Complete step-by-step answer:
Let tan⁡−1(1+2x1−2x){\tan ^{ - 1}}(\dfrac{{1 + 2x}}{{1 - 2x}}) be u and 1+4x2\sqrt {1 + 4{x^2}} be v.
Now, let’s find dudx\dfrac{{du}}{{dx}}and dvdx\dfrac{{dv}}{{dx}} separately.
We know that, any inverse function of the form tan⁡−1(A+BA−B)=tan⁡−1A+tan⁡−1B{\tan ^{ - 1}}(\dfrac{{A + B}}{{A - B}}) = {\tan ^{ - 1}}A + {\tan ^{ - 1}}B.
Now, in this case tan⁡−1(1+2x1−2x){\tan ^{ - 1}}(\dfrac{{1 + 2x}}{{1 - 2x}}) can be written as tan⁡−1(1)+tan⁡−1(2x){\tan ^{ - 1}}(1) + {\tan ^{ - 1}}(2x)
So, dudx\dfrac{{du}}{{dx}} = ddx\dfrac{d}{{dx}}(tan⁡−1(1)+tan⁡−1(2x){\tan ^{ - 1}}(1) + {\tan ^{ - 1}}(2x))
Now we also know that tan⁡−1x=11+x2{\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}
so, dudx\dfrac{{du}}{{dx}} = (0+ddx\dfrac{d}{{dx}} (11+x2)(\dfrac{1}{{1 + {x^2}}}))
dudx\dfrac{{du}}{{dx}}=21+4x2\dfrac{2}{{1 + 4{x^2}}} - equation (1)
So, we have the value of dudx\dfrac{{du}}{{dx}}, now we need the value of dvdx\dfrac{{dv}}{{dx}}=ddx(1+4x2)\dfrac{d}{{dx}}(\sqrt {1 + 4{x^2}} )
now any function of the form d(f)dx=12f\dfrac{{d(\sqrt f )}}{{dx}} = \dfrac{1}{{2\sqrt f }}
so, dvdx\dfrac{{dv}}{{dx}}=12×11+4x2(0+8x)\dfrac{1}{2} \times \dfrac{1}{{\sqrt {1 + 4{x^2}} }}(0 + 8x)
dvdx\dfrac{{dv}}{{dx}}=4x1+4x2\dfrac{{4x}}{{\sqrt {1 + 4{x^2}} }} -equation (2)
Now to find dudv\dfrac{{du}}{{dv}} we just need to divide equation (1) from equation (2)
dudv\dfrac{{du}}{{dv}}=21+4x2×1+4x24x\dfrac{2}{{1 + 4{x^2}}} \times \dfrac{{\sqrt {1 + 4{x^2}} }}{{4x}}
dudv\dfrac{{du}}{{dv}}=12x1+4x2\dfrac{1}{{2x\sqrt {1 + 4{x^2}} }}

Note: In questions where you have to differentiate trigonometric functions, identities and formulas are always the primary key. It helps you to simplify the solution, for example in this question we used three different properties that are
1)d(f)dx=12f\dfrac{{d(\sqrt f )}}{{dx}} = \dfrac{1}{{2\sqrt f }}
2)tan⁡−1(A+BA−B)=tan⁡−1A+tan⁡−1B{\tan ^{ - 1}}(\dfrac{{A + B}}{{A - B}}) = {\tan ^{ - 1}}A + {\tan ^{ - 1}}B
3)tan⁡−1x=11+x2{\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}
so, try to remember as many properties and identities as possible.