If $y = x\sqrt{a^2 + x^2} + a^2\log(x + \sqrt{a^2 + x^2})$, then show that $\frac{dy}{dx} = 2\sqrt{a^2 + x^2}$.

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Accepted Answer

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> [!TLDR]
> Differentiate the two terms separately — the first with the product rule, the second using the standard derivative of $\log(x+\sqrt{a^2+x^2})$ — then add and simplify.
>
> $$y = x\sqrt{a^2+x^2} + a^2\log\!\left(x+\sqrt{a^2+x^2}\right)$$
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> **Term 1:**
> $$\frac{d}{dx}\left[x\sqrt{a^2+x^2}\right] = \sqrt{a^2+x^2} + x\cdot\frac{x}{\sqrt{a^2+x^2}} = \frac{(a^2+x^2)+x^2}{\sqrt{a^2+x^2}} = \frac{a^2+2x^2}{\sqrt{a^2+x^2}}$$
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> **Term 2:**
> $$\frac{d}{dx}\left[a^2\log\!\left(x+\sqrt{a^2+x^2}\right)\right] = a^2\cdot\frac{1+\dfrac{x}{\sqrt{a^2+x^2}}}{x+\sqrt{a^2+x^2}} = a^2\cdot\frac{\left(\sqrt{a^2+x^2}+x\right)/\sqrt{a^2+x^2}}{x+\sqrt{a^2+x^2}} = \frac{a^2}{\sqrt{a^2+x^2}}$$
> …

Q.If $y = x\sqrt{a^2 + x^2} + a^2\log(x + \sqrt{a^2 + x^2})$ , then show that $\frac{dy}{dx} = 2\sqrt{a^2 + x^2}$ .

Concept understanding — Chain Rule

The Chain Rule: Why It Makes Sense

The Precise Statement

A Simple Example

Why It's Called a "Chain"

Q.If y=xa2+x2+a2log⁡(x+a2+x2)y = x\sqrt{a^2 + x^2} + a^2\log(x + \sqrt{a^2 + x^2})y=xa2+x2​+a2log(x+a2+x2​), then show that dydx=2a2+x2\frac{dy}{dx} = 2\sqrt{a^2 + x^2}dxdy​=2a2+x2​.

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Q.If $y = x\sqrt{a^2 + x^2} + a^2\log(x + \sqrt{a^2 + x^2})$ , then show that $\frac{dy}{dx} = 2\sqrt{a^2 + x^2}$ .

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