Question

Differentiate the following :

If $y = \sqrt{{x}^{2}+1}-log \bigg(\dfrac{1}{x} + \sqrt{x + \dfrac{1}{{x}^{2}}}\bigg)$

Answer 1

$\mathsf{\frac{dy}{dx}=\frac{x}{\sqrt{x^{2}+1}}-\frac{3}{2}.\frac{x^{2}}{(1+\sqrt{x^{3}+1})\sqrt{x^{3}+1}}+\frac{1}{x}}$

Step-by-step explanation:

Let us differentiate each term separately and it will make the understandings easier.

1st term = $\mathsf{\sqrt{x^{2}+1}}$

             = $\mathsf{(x^{2}+1)^{\frac{1}{2}}}$

Differentiating with respect to x, we get

  $\mathsf{\frac{d}{dx}(\sqrt{x^{2}+1})}$

$\mathsf{=\frac{d}{dx}(x^{2}+1)^{\frac{1}{2}}}$

$\mathsf{=\frac{1}{2}.(x^{2}+1)^{\frac{1}{2}-1}.\frac{d}{dx}(x^{2}+1)}$

$\mathsf{=\frac{1}{2}.(x^{2}+1)^{-\frac{1}{2}}.2x}$

$\mathsf{=\frac{x}{\sqrt{x^{2}+1}}}$

2nd term = $\mathsf{log\big(\frac{1}{x}+\sqrt{x+\frac{1}{x^{2}}}\big)}$

               = $\mathsf{log\big(\frac{1}{x}+\frac{1}{x}\sqrt{x^{3}+1}\big)}$

               = $\mathsf{log\{\frac{1}{x}(1+\sqrt{x^{3}+1})\}}$

               = $\mathsf{log(1+\sqrt{x^{3}+1})-logx}$

Differentiating with respect to x, we get

  $\mathsf{\frac{d}{dx}\big[log\big(\frac{1}{x}+\sqrt{x+\frac{1}{x^{3}}}\big)\big]}$

$\mathsf{=\frac{d}{dx}\{log(1+\sqrt{x^{3}+1})-logx\}}$

$\mathsf{=\frac{d}{dx}\{log(1+\sqrt{x^{3}+1})\}-\frac{d}{dx}(logx)}$

$\mathsf{=\frac{1}{1+\sqrt{x^{3}+1}}.\frac{d}{dx}(1+\sqrt{x^{3}+1})-\frac{1}{x}}$

$\mathsf{=\frac{1}{1+\sqrt{x^{3}+1}}.\frac{d}{dx}(x^{3}+1)^{\frac{1}{2}}-\frac{1}{x}}$

$\mathsf{=\frac{1}{1+\sqrt{x^{3}+1}}.\frac{1}{2}.(x^{3}+1)^{\frac{1}{2}-1}.\frac{d}{dx}(x^{3}+1)-\frac{1}{x}}$

$\mathsf{=\frac{1}{1+\sqrt{x^{3}+1}}.\frac{1}{2}.(x^{3}+1)^{-\frac{1}{2}}.3x^{2}-\frac{1}{x}}$

$\mathsf{=\frac{3}{2}.\frac{x^{2}}{(1+\sqrt{x^{3}+1})\sqrt{x^{3}+1}}-\frac{1}{x}}$

Now we put the differentiated values of the first and the second term in the differentiation of y with respect to x, i.e.,

$\mathsf{\frac{dy}{dx}=\frac{x}{\sqrt{x^{2}+1}}-\frac{3}{2}.\frac{x^{2}}{(1+\sqrt{x^{3}+1})\sqrt{x^{3}+1}}+\frac{1}{x}}$

Answer 2

Answer:

$\tan(96) \pi \gamma \cot(87)$

$this \: is \: the \: formulae$

$hope \: it \: helps \: you$