How do you differentiate $f (x) = e^{\sqrt{x^{2} + 2}}$ $f(x)=e^(sqrt(x^2+2))$ ?

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How do you differentiate $f (x) = e^{\sqrt{x^{2} + 2}}$ $f(x)=e^(sqrt(x^2+2))$ ?

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Answer 1 · 2018-03-15T13:02:45

$\frac{x e^{\sqrt{x^{2} + 2}}}{\sqrt{x^{2} + 2}}$ $(xe^(sqrt(x^2+2)))/(sqrt(x^2+2))$

Explanation:

We use the chain rule, which states that

$\frac{d f}{d x} = \frac{d f}{d u} \cdot \frac{d u}{d x}$ $(df)/dx=(df)/(du)*(du)/dx$

Let $u = \sqrt{x^{2} + 2}$ $u=sqrt(x^2+2)$ . Now we find differentiate $\sqrt{x^{2} + 2}$ $sqrt(x^2+2)$ to find $\frac{d u}{d x}$ $(du)/dx$ .

We use the chain rule again, and so $t = x^{2} + 2, \frac{d t}{d x} = 2 x$ $t=x^2+2, (dt)/dx=2x$ , $f = \sqrt{t}, \frac{d f}{d t} = \frac{1}{2 \sqrt{t}}$ $f=sqrt(t), (df)/(dt)=1/(2sqrt(t)$ .

Then, $\frac{d f}{d x} = 2 x \cdot \frac{1}{2 \sqrt{t}}$ $(df)/dx=2x*1/(2sqrt(t)$

$= \frac{x}{\sqrt{t}}$ $=x/(sqrt(t)$

Reversing the substitution that $t = x^{2} + 2$ $t=x^2+2$ , we get

$= \frac{x}{\sqrt{x^{2} + 2}}$ $=x/sqrt(x^2+2)$ or $\frac{d u}{d x} = \frac{x}{\sqrt{x^{2} + 2}}$ $(du)/dx=x/sqrt(x^2+2)$

Now, we got $f = e^{u}, \frac{d f}{d u} = e^{u}$ $f=e^u, (df)/(du)=e^u$ . Combining together, we get

$\frac{d f}{d x} = e^{u} \cdot \frac{x}{\sqrt{x^{2} + 2}}$ $(df)/dx=e^u*x/sqrt(x^2+2)$

$= \frac{x e^{u}}{\sqrt{x^{2} + 2}}$ $=(xe^u)/sqrt(x^2+2)$

Reversing the substitution that $u = \sqrt{x^{2} + 2}$ $u=sqrt(x^2+2)$ , we have

$= \frac{x e^{\sqrt{x^{2} + 2}}}{\sqrt{x^{2} + 2}}$ $=(xe^(sqrt(x^2+2)))/(sqrt(x^2+2))$

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How do you differentiate $f (x) = e^{\sqrt{x^{2} + 2}}$ $f(x)=e^(sqrt(x^2+2))$ ?

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