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  • Differentiate each of the following w.r.t. x log(x+x^+a)

Differentiate each of the following w.r.t. x
\log\left ( x+\sqrt{x^2 +a}\right )

 

Answers (1)

Given:
\log \left(x+\sqrt{x^{2}+a}\right)

\mathrm{y}=\log \mathrm{t}
Now, Differentiate w.r.t t
\\\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\log \mathrm{t})$ \\$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{1}{\mathrm{t}}
And, t=x+\sqrt{x^{2}+a}
Now, differentiate w.r.t x
\\\frac{d t}{d x}=\frac{d}{d x}\left(x+\sqrt{x^{2}+a}\right)$ \\$\frac{d t}{d x}=\left[1+\frac{d}{d x}\left(\left(x^{2}+a\right)^{\frac{1}{2}}\right)\right.

\\\frac{d t}{d x}=\left[1+\frac{1}{2}\left(x^{2}+a\right)^{-\frac{1}{2}} \cdot \frac{d}{d x}\left(x^{2}\right)\right]$ \\\\$\frac{d t}{d x}=\left[1+\frac{1}{2}\left(x^{2}+a\right)^{-\frac{1}{2}} \cdot(2 x)\right]
Now, using chain rule, we get
\\\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}$ \\\\$\frac{d y}{d x}=\frac{1}{t} \times\left[1+\frac{1}{2}\left(x^{2}+a\right)^{-\frac{1}{2}} \cdot(2 x)\right]
Substitute the value of t

\frac{d y}{d x}=\frac{1}{x+\sqrt{x^{2}+a}} \times\left[1+\frac{1}{2}\left(x^{2}+a\right)^{-\frac{1}{2}} \cdot(2 x)\right]$ \\$=\frac{1}{x+\sqrt{x^{2}+a}} \times\left[1+\frac{x}{ \sqrt{x^{2}+a}}\right]$ \\$=\frac{\sqrt{x^{2}+a}+x}{\left(x+\sqrt{x^{2}+a}\right) \sqrt{x^{2}+a}}

 =\frac{1}{\sqrt{x^{2}+a}} \\
 

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