If y-x-x2-a2n- then dy-dx-

The correct option is A ny√((x^2+a^2)) Given, y={x+√((x^2+a^2))}^n ⇒ #10; #10;dy/dx=n[x+√((x^2+a^2))]^n 1·{ #10;1+1/2√(x^2+a^2)..2x } ...

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Chain Rule of Differentiation

If y=x+√x2+...

Question

If $y = {x + \sqrt{(x^{2} + a^{2})}}^{n},$ then $\frac{d y}{d x} =$

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

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

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\frac{(n - 1) x}{\sqrt{(x^{2} + a^{2})}}

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\frac{(n - 1) y}{\sqrt{(x^{2} + a^{2})}}

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Solution

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Similar questions

If $y = {[x + \sqrt{x^{2} + a^{2}}]}^{n}, t h e n \frac{d y}{d x}$ is equal to

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

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

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}})

\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})}}}

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