Question

If $$y=\tan^{-1} \left( \dfrac{ax-b}{bx+a} \right) $$, prove that $$\dfrac{dy}{dx} =\dfrac{1}{\left( 1+x^{2}\right) }$$

Answer 1

Hint: In this question it is given that if $$y=\tan^{-1} \left( \dfrac{ax-b}{bx+a} \right) $$,then we have to prove that $$\dfrac{dy}{dx} =\dfrac{1}{\left( 1+x^{2}\right) }$$.
So to find the solution we have to simply differentiate the given function with respect to ‘x’ and during integration we have to use some formulas, which are,
1) If f(x) be the the function of x, then, $$\dfrac{d}{dx} \tan^{-1} \left( f\left( x\right) \right) =\dfrac{1}{1+\left( f\left( x\right) \right)^{2} } \cdot \dfrac{d}{dx} \left( f\left( x\right) \right) $$
2) If u and v are the the function of x, then, $$\dfrac{d}{dx} \left( \dfrac{u}{v} \right) =\dfrac{v\dfrac{du}{dx} -u\dfrac{dv}{dx} }{v^{2}}$$
Complete step-by-step solution:
Given,
$$y=\tan^{-1} \left( \dfrac{ax-b}{bx+a} \right) $$
Now differentiating both side w.r.t ‘x’, we get,
$$\dfrac{dy}{dx} =\dfrac{d}{dx} \left( \tan^{-1} \left( \dfrac{ax-b}{bx+a} \right) \right) $$
$$=\dfrac{1}{1+\left( \dfrac{ax-b}{bx+a} \right)^{2} } \cdot \dfrac{d}{dx} \left( \dfrac{ax-b}{bx+a} \right) $$ [by using formula (1)]
$$=\dfrac{\left( bx+a\right)^{2} }{\left( bx+a\right)^{2} +\left( ax-b\right)^{2} } \cdot \dfrac{\left( bx+a\right) \cdot \dfrac{d}{dx} \left( ax-b\right) -\left( ax-b\right) \cdot \dfrac{d}{dx} \left( bx+a\right) }{\left( bx+a\right)^{2} }$$ [ by using formula (2)]
$$=\dfrac{\left( bx+a\right)^{2} }{\left( bx+a\right)^{2} +\left( ax-b\right)^{2} } \cdot \dfrac{\left( bx+a\right) \cdot a-\left( ax-b\right) \cdot b}{\left( bx+a\right)^{2} }$$
$$=\dfrac{a\left( bx+a\right) -b\left( ax-b\right) }{\left( bx+a\right)^{2} +\left( ax-b\right)^{2} }$$
$$=\dfrac{abx+a^{2}-abx+b^{2}}{\left( bx+a\right)^{2} +\left( ax-b\right)^{2} }$$
Now as we know that $$\left( p+q\right)^{2} =p^{2}+2pq+q^{2}$$ and $$\left( p-q\right)^{2} =p^{2}-2pq+q^{2}$$, so by these identity we can write the above equation as,
$$\dfrac{dy}{dx}=\dfrac{a^{2}+b^{2}}{b^{2}x^{2}+2abx+a^{{}2}+a^{2}x^{2}-2abx+b^{2}}$$
$$=\dfrac{a^{2}+b^{2}}{b^{2}x^{2}+a^{2}x^{2}+b^{2}+a^{2}}$$
$$=\dfrac{(b^{2}+a^{2})}{(b^{2}+a^{2})x^{2}+(b^{2}+a^{2})}$$
$$=\dfrac{(b^{2}+a^{2})}{(b^{2}+a^{2})\left( x^{2}+1\right) }$$ [taking $a^{2}+b^{2}$ from the denominator]
$$=\dfrac{1}{\left( x^{2}+1\right) }$$
$$=\dfrac{1}{\left( 1+x^{2}\right) }$$
Therefore, we have,
$$\dfrac{dy}{dx} =\dfrac{1}{\left( 1+x^{2}\right) }$$
Hence proved.
Note: While proving this kind of question you need to know the basic formulas of derivative that we have already mentioned in the hint portion also derivative of any polynomial, $$ax^{2}\pm bx\pm c$$ is can be written as,
$$\dfrac{d}{dx} \left( ax^{2}\pm bx\pm c\right)$$
$$=\dfrac{d}{dx} \left( ax^{2}\right) \pm \dfrac{d}{dx} \left( bx\right) \pm \dfrac{d}{dx} \left( c\right) $$
$$=a\dfrac{d}{dx} \left( x^{2}\right) \pm b$$
Where a,b and c are the constant and the derivative of a constant term is 0 and if any constant is multiplied with a variable then the derivative of that constant term can be taken outside.