Question

Find d/dx tan⁻¹ 4x/1+21x², x > 0

Answer 1

we have to find the value of $\frac{d}{dx}tan^{-1}\frac{4x}{1+21x^2}$

we know, differentiation of $tan^{-1}x=\frac{1}{1+x^2}$

now, $\frac{d}{dx}tan^{-1}\frac{4x}{1+21x^2}=\frac{1}{1+\{\frac{4x}{1+21x^2}\}^2}\times\frac{d\{\frac{4x}{1+21x^2}\}}{dx}\\\\=\frac{(1+21x^2)^2}{1+441x^4+42x^2+16x^2}\frac{(1+21x^2)4-4x(42x)}{(1+21x^2)^2}\\\\=\frac{(1+21x^2)^2}{1+(21x^2)^2+58x^2}\frac{4(1-21x^2)}{(1+21x^2)^2}\\\\=\frac{4(1-21x^2)}{1+(21x^2)^2+58x^2}$

Answer 2

Answer:-

See attached image