Question

Derivative of tan^-1 ((a+x)/(1-ax)) wrt (4x^3 - 3x)

Answer 1

Let,

$\longrightarrow u=\tan^{-1}\left(\dfrac{1+2x}{1-2x}\right)\quad\quad\dots(1)$

Substitute,

$\longrightarrow v=\sqrt{1+4x}$

Thus the derivative of $\tan^{-1}\left(\dfrac{1+2x}{1-2x}\right)tan−1(1−2x1+2x) with \:respect \:to \sqrt{1+4x}1+4x will\: be \dfrac{du}{dv}$

$\longrightarrow v^2=1+4x$

$\longrightarrow x=\dfrac{v^2-1}{4}$

Then (1) becomes,

$\longrightarrow u=\tan^{-1}\left(\dfrac{1+2\cdot\frac{v^2-1}{4}}{1-2\cdot\frac{v^2-1}{4}}\right)$

$\longrightarrow u=\tan^{-1}\left(\dfrac{1+\frac{v^2-1}{2}}{1-\frac{v^2-1}{2}}\right)$

$\longrightarrow u=\tan^{-1}\left(\dfrac{2+v^2-1}{2-v^2+1}\right)$

$\longrightarrow u=\tan^{-1}\left(\dfrac{v^2+1}{3-v^2}\right)\quad\quad\dots(2)$

Let,

$\longrightarrow w=\dfrac{v^2+1}{3-v^2}\quad\quad\dots(3)$

Differentiating wrt v,v,

$\longrightarrow \dfrac{dw}{dv}=\dfrac{d}{dv}\left(\dfrac{v^2+1}{3-v^2}\right)$

By quotient rule,

$\longrightarrow \dfrac{dw}{dv}=\dfrac{2v(3-v^2)+2v(v^2+1)}{(3-v^2)^2}$

$\longrightarrow \dfrac{dw}{dv}=\dfrac{2v(3-v^2+v^2+1)}{(3-v^2)^2}$

$\longrightarrow \dfrac{dw}{dv}=\dfrac{8v}{(3-v^2)^2}\quad\quad\dots(4)$

Then (2) becomes,

$\longrightarrow u=\tan^{-1}\left(w\right)$

Differentiating wrt v,v,

$\longrightarrow \dfrac{du}{dv}=\dfrac{d}{dv}\big(\tan^{-1}\left(w\right)\big)$

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Answer 2

Answer:

⟶u=tan

−1

(

1−2x

1+2x

)…(1)

Substitute,

\longrightarrow v=\sqrt{1+4x}⟶v=

1+4x

Thus the derivative of \tan^{-1}\left(\dfrac{1+2x}{1-2x}\right)tan−1(1−2x1+2x) with \:respect \:to \sqrt{1+4x}1+4x will\: be \dfrac{du}{dv}tan

−1

(

1−2x

1+2x

)tan−1(1−2x1+2x) withrespectto

1+4x

1+4x willbe

dv

du

\longrightarrow v^2=1+4x⟶v

2

=1+4x

\longrightarrow x=\dfrac{v^2-1}{4}⟶x=

4

v

2

−1

Then (1) becomes,

\longrightarrow u=\tan^{-1}\left(\dfrac{1+2\cdot\frac{v^2-1}{4}}{1-2\cdot\frac{v^2-1}{4}}\right)⟶u=tan

−1

(

1−2⋅

4

v

2

−1

1+2⋅

4

v

2

−1

)

\longrightarrow u=\tan^{-1}\left(\dfrac{1+\frac{v^2-1}{2}}{1-\frac{v^2-1}{2}}\right)⟶u=tan

−1

(

1−

2

v

2

−1

1+

2

v

2

−1

)

\longrightarrow u=\tan^{-1}\left(\dfrac{2+v^2-1}{2-v^2+1}\right)⟶u=tan

−1

(

2−v

2

+1

2+v

2

−1

)

\longrightarrow u=\tan^{-1}\left(\dfrac{v^2+1}{3-v^2}\right)\quad\quad\dots(2)⟶u=tan

−1

(

3−v

2

v

2

+1

)…(2)

Let,

\longrightarrow w=\dfrac{v^2+1}{3-v^2}\quad\quad\dots(3)⟶w=

3−v

2

v

2

+1

…(3)

Differentiating wrt v,v,

\longrightarrow \dfrac{dw}{dv}=\dfrac{d}{dv}\left(\dfrac{v^2+1}{3-v^2}\right)⟶

dv

dw

=

dv

d

(

3−v

2

v

2

+1

)

By quotient rule,

\longrightarrow \dfrac{dw}{dv}=\dfrac{2v(3-v^2)+2v(v^2+1)}{(3-v^2)^2}⟶

dv

dw

=

(3−v

2

)

2

2v(3−v

2

)+2v(v

2

+1)

\longrightarrow \dfrac{dw}{dv}=\dfrac{2v(3-v^2+v^2+1)}{(3-v^2)^2}⟶

dv

dw

=

(3−v

2

)

2

2v(3−v

2

+v

2

+1)

\longrightarrow \dfrac{dw}{dv}=\dfrac{8v}{(3-v^2)^2}\quad\quad\dots(4)⟶

dv

dw

=

(3−v

2

)

2

8v

…(4)

Then (2) becomes,

\longrightarrow u=\tan^{-1}\left(w\right)⟶u=tan

−1

(w)

Differentiating wrt v,v,

\longrightarrow \dfrac{du}{dv}=\dfrac{d}{dv}\big(\tan^{-1}\left(w\right)\big)⟶

dv

du

=

dv

d

(tan

−1

(w))