If y-x-ex- then d2ydx2 -

y=x+ex Differentiating both sides with respect to x, we get dydx=ddxx+ex #8658;dydx=1+ex nbsp; nbsp; Again differentiating both sides with respect t ...

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Standard IX

Mathematics

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If y=x+ex, th...

Question

If y = x + e^x, then $\frac{d^{2} y}{d x^{2}}$ = _____________________.

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Solution

$y = x + e^{x}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} (x + e^{x})$

$\Rightarrow \frac{d y}{d x} = 1 + e^{x}$

Again differentiating both sides with respect to x, we get

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (1 + e^{x})$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = 0 + e^{x} = e^{x}$

$∴ \frac{d^{2} y}{d x^{2}} = e^{x}$

If y = x + e^x, then $\frac{d^{2} y}{d x^{2}}$ = $e^{x}$ .

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If y = x + ex, then d2ydx2 = _____________________.

y=x+ex Differentiating both sides with respect to x, we get dydx=ddxx+ex ⇒dydx=1+ex Again differentiating both sides with respect to x, we get ddxdydx=ddx1+ex ⇒d2ydx2=0+ex=ex ∴d2ydx2=ex If y = x + ex, then d2ydx2 = ex .

If y = x + e^x, then $\frac{d^{2} y}{d x^{2}}$ = _____________________.