Given y- x 2 - a 2 - dy dx -

The correct option is D x √( x ^ 2 + a ^ 2 ) Given y= √(x^2+a^2) ⇒dydx= 12×1√(x^2+a^2)× 2x=x√(x^2+a^2) ...

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Chain Rule of Differentiation

Given y=√ x...

Question

Given $y = \sqrt{x^{2} + a^{2}}, \frac{d y}{d x} =$

\frac{x}{2 \sqrt{x^{2} + a^{2}}}

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\frac{2 x}{\sqrt{x^{2} + a^{2}}}

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\frac{x}{\sqrt{x^{2} + a^{2}}}

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2 x \sqrt{x^{2} + a^{2}}

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Solution

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Similar questions

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

\frac{d y}{d x} = 2 \sqrt{a^{2} + x^{2}}

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

\frac{d y}{d x} = - \frac{2 a^{2}}{x^{3}} {1 + \frac{a^{2}}{\sqrt{(a^{4} - x^{4})}}}

y = [x + \sqrt{x^{2} + a^{2}}]^{n}

\frac{d y}{d x} = \frac{n y}{\sqrt{x^{2} + a^{2}}}

y = x \sqrt{a^{2} + x^{2}} + a^{2} log (x + \sqrt{a^{2} + x^{2}}) t h e n \frac{d y}{d x} = 2 \sqrt{a^{2} + x^{2}}

\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = l n | x + \sqrt{x^{2} + a^{2}} | + C

\int \frac{d x}{\sqrt{x^{2} - a^{2}}} = l n | x - \sqrt{x^{2} - a^{2}} | + C

\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} l n ∣ ∣ \frac{x - a}{x + a} ∣ ∣ + C

\int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} l n ∣ ∣ \frac{x + a}{x - a} ∣ ∣ + C

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