How do you differentiate $f (x) = e^{\sqrt{3 ln x + x^{2}}}$ $f(x)=e^sqrt(3lnx+x^2)$ using the chain rule.?

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How do you differentiate $f (x) = e^{\sqrt{3 ln x + x^{2}}}$ $f(x)=e^sqrt(3lnx+x^2)$ using the chain rule.?

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Answer 1 · 2016-01-16T21:38:59

$\frac{d y}{d x} = \frac{e^{\sqrt{3 ln x + x^{2}}} \cdot (\frac{3}{x} + 2 x)}{(2 \sqrt{3 ln x + x^{2}})}$ $dy/dx=(e^sqrt(3lnx+x^2) * (3/x+2x))/((2sqrt(3lnx+x^2))$

Explanation:

The chain rule:

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d v} \cdot \frac{d v}{d x}$ $dy/dx=dy/(du) * (du)/(dv)*(dv)/(dx)$

$y = e^{u}, \frac{d y}{d u} = e^{u}$ $y = e^u, dy/(du)=e^u$

$u = v^{\frac{1}{2}}, \frac{d u}{d v} = \frac{1}{2} v^{- (\frac{1}{2})}$ $u=v^(1/2), (du)/(dv)=1/2v^-(1/2)$

$v = 3 ln x + x^{2}, \frac{d v}{d x} = \frac{3}{x} + 2 x$ $v=3lnx+x^2, (dv)/dx=3/x+2x$

$\frac{d y}{d x} = e^{u} \cdot \frac{1}{2 \sqrt{v}} \cdot (\frac{3}{x} + 2 x)$ $dy/dx=e^u*1/(2sqrtv)*(3/x+2x)$

$\frac{d y}{d x} = \frac{e^{\sqrt{3 ln x + x^{2}}} \cdot (\frac{3}{x} + 2 x)}{(2 \sqrt{3 ln x + x^{2}})}$ $dy/dx=(e^sqrt(3lnx+x^2) * (3/x+2x))/((2sqrt(3lnx+x^2))$

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How do you differentiate $f (x) = e^{\sqrt{3 ln x + x^{2}}}$ $f(x)=e^sqrt(3lnx+x^2)$ using the chain rule.?

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Explanation:

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How do you differentiate f(x)=e^sqrt(3lnx+x^2)f(x)=e√3lnx+x2 f(x)=e^sqrt(3lnx+x^2) using the chain rule.?

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How do you differentiate $f (x) = e^{\sqrt{3 ln x + x^{2}}}$ $f(x)=e^sqrt(3lnx+x^2)$ using the chain rule.?