If y-x-x2-a2-n- then prove that dydx-ny-x2-a2

Given, y=(x+√(x^2+a^2))^n differentiating on both sides, we get, dydx=n(x+√(x^2+a^2))^n 1(1+x√(x^2+a^2)) =n(x+√(x^2+a^2))^n 1(√(x^2+a^2)+x√(x^2+a^2 ...

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Byju's Answer

Standard XII

Mathematics

Monotonically Increasing Functions

If y=[x+√x2...

Question

If $y = [x + \sqrt{x^{2} + a^{2}}]^{n}$ , then prove that $\frac{d y}{d x} = \frac{n y}{\sqrt{x^{2} + a^{2}}}$

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Solution

Given,

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

differentiating on both sides, we get,

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

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

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

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

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

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If y=[x+√x2+a2]n, then prove that dydx=ny√x2+a2

Given,y=(x+√x2+a2)ndifferentiating on both sides, we get,dydx=n(x+√x2+a2)n−1(1+x√x2+a2)=n(x+√x2+a2)n−1(√x2+a2+x√x2+a2)=n(x+√x2+a2)n−1+1√x2+a2=n(x+√x2+a2)n√x2+a2∴dydx=ny√x2+a2

If $y = [x + \sqrt{x^{2} + a^{2}}]^{n}$ , then prove that $\frac{d y}{d x} = \frac{n y}{\sqrt{x^{2} + a^{2}}}$