Name: Differentiation of Exponential Functions
Uploaded: 2019-07-19
Duration: 12 min 3 s
Description: To know how to find the derivative of exponential functions or multiple functions containing the exponential term. Let's watch this video and explore!

Question

If

y = \log (x + \sqrt{a^{2} + x^{2}})

, show that

(a^{2} + x^{2}) y_{2} + x y_{1} = 0 .

Accepted Answer

The given function is,

$y = \log (x + \sqrt{a^{2} + x^{2}}) \dots \dots \dots (1)$

Differentiating both sides of equation $(1)$ w.r.t. $x$ , we get

$\frac{d}{d x} (y) = \frac{d}{d x} [\log (x + \sqrt{a^{2} + x^{2}})]$

$y_{1} = \frac{1}{(x + \sqrt{a^{2} + x^{2}})} \times \frac{d}{d x} (x + \sqrt{a^{2} + x^{2}})$ $[$ Use chain rule and also $\frac{d}{d x} (\log x) = \frac{1}{x}]$

$y_{1} = \frac{1}{(x + \sqrt{a^{2} + x^{2}})} \times [\frac{d}{d x} (x) + \frac{d}{d x} (\sqrt{a^{2} + x^{2}})]$ $[∵ \frac{d}{d x} [f (x) + g (x)] = \frac{d}{d x} f (x) + \frac{d}{d x} g (x)]$

$y_{1} = \frac{1}{(x + \sqrt{a^{2} + x^{2}})} \times [1 + \frac{1}{2 \sqrt{x^{2} + 1}} \frac{d}{d x} (a^{2} + x^{2})]$ $[∵ \frac{d}{d x} (\sqrt{x}) = \frac{1}{2 \sqrt{x}}]$

$y_{1} = \frac{1}{(x + \sqrt{a^{2} + x^{2}})} \times [1 + \frac{1}{2 \sqrt{a^{2} + x^{2}}} [\frac{d}{d x} (a^{2}) + (x^{2})]]$ $[∵ \frac{d}{d x} [f (x) + g (x)] = \frac{d}{d x} f (x) + \frac{d}{d x} g (x)]$

$y_{1} = \frac{1}{(x + \sqrt{a^{2} + x^{2}})} \times [1 + \frac{1}{2 \sqrt{a^{2} + x^{2}}} [0 + 2 x]]$ $[∵ \frac{d}{d x} x^{n} = n x^{n - 1}]$

$y_{1} = [\frac{1}{x + \sqrt{a^{2} + x^{2}}}] [1 + \frac{1 \times 2 x}{2 \sqrt{a^{2} + x^{2}}}]$

$y_{1} = [\frac{1}{x + \sqrt{a^{2} + x^{2}}}] [\frac{\sqrt{a^{2} + x^{2}} + x}{\sqrt{a^{2} + x^{2}}}]$

$y_{1} = \frac{1}{\sqrt{a^{2} + x^{2}}} = {(a^{2} + x^{2})}^{- \frac{1}{2}} \dots \dots \dots (2)$

Again differentiating both sides of equation $(2)$ w.r.t. $x$ , we get

$\frac{d}{d x} (y_{1}) = \frac{d}{d x} ({(a^{2} + x^{2})}^{- \frac{1}{2}})$

$y_{2} = - \frac{1}{2} {(a^{2} + x^{2})}^{- \frac{1}{2} - 1} \frac{d}{d x} (a^{2} + x^{2})$ $[$ Use chain rule and also $\frac{d}{d x} x^{n} = n x^{n - 1}]$

$y_{2} = - \frac{1}{2} {(a^{2} + x^{2})}^{- \frac{3}{2}} [\frac{d}{d x} (a^{2}) + \frac{d}{d x} (x^{2})]$ $[∵ \frac{d}{d x} [f (x) + g (x)] = \frac{d}{d x} f (x) + \frac{d}{d x} g (x)]$

$y_{2} = - \frac{1}{2} \frac{1}{{(a^{2} + x^{2})}^{\frac{3}{2}}} [0 + 2 x]$ $[∵ \frac{d}{d x} x^{n} = n x^{n - 1}]$

$y_{2} = - \frac{1}{2} \frac{2 x}{(a^{2} + x^{2}) \sqrt{a^{2} + x^{2}}}$

$(a^{2} + x^{2}) y_{2} = - \frac{x}{\sqrt{a^{2} + x^{2}}}$ $[$ Cross multiply $(x^{2} + 1)]$

$(a^{2} + x^{2}) y_{2} = - y_{1} x$ $[$ Substituting value from equation $(2)]$

$(a^{2} + x^{2}) y_{2} + y_{1} x = 0$ .

Hence proved.

If $y = \log (x + \sqrt{a^{2} + x^{2}})$ , show that $(a^{2} + x^{2}) y_{2} + x y_{1} = 0 .$

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

If y=logx+a2+x2, show that a2+x2y2+xy1=0.

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

If $y = \log (x + \sqrt{a^{2} + x^{2}})$ , show that $(a^{2} + x^{2}) y_{2} + x y_{1} = 0 .$