Question

Solve the following
Differentiate w.r.t. x:
1. ㏒(x + √(x² + a²)
2.㏒(√(x-a) + √(x-b) )

Answer 1

Answer:

Step-by-step explanation:

1. $log (x + \sqrt{x^2 + a^2})$

Solution:

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

$\frac{dy}{dx} = \frac{1}{x + \sqrt{(x^2+a^2)} } * 1+\frac{1}{2\sqrt{(x^2+a^2)} }* 2x$

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

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

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

Derivative of $log (x + \sqrt{x^2 + a^2})$ is $\frac{1}{\sqrt{(x^2+a^2)} }$

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Formula used

$\frac{d}{dx}(log x) = \frac{1}{x}$

$\frac{d}{dx}(\sqrt{x} ) = \frac{1}{2\sqrt{x} }$

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2. $log(\sqrt{(x-a)} + \sqrt{(x - b)}$

$\frac{d}{dx} = \frac{1}{\sqrt{(x-a)}+\sqrt{(x-b)} } * \frac{1}{2\sqrt{(x- a)} } +\frac{1}{2\sqrt{x} (x-b)}$

$= \frac{\sqrt{(x}-b) +\sqrt{(x-a)} }{\sqrt{(x-a)}+\sqrt{(x-b)}2\sqrt{(x-a)}\sqrt{(x-b)} }$

$= \frac{1}{2\sqrt{(x-a)}\sqrt{(x-b)} }$

Derivative of $log(\sqrt{(x-a)} + \sqrt{(x - b)}$ is $= \frac{1}{2\sqrt{(x-a)}\sqrt{(x-b)} }$

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Formula used

$\frac{d}{dx}(log x) = \frac{1}{x}$

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More details

1. f(x) = f'(x)
2. $x^n = nx^{n-1}$
3. $e^{ax} = ae^{ax}$
4. ㏒ x = $\frac{1}{x}$
5. Constant = 0
6. $\sqrt{x} = \frac{1}{2\sqrt{x} }$