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October 5th, 2016, 08:11 AM   #1
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combinding chain and power rule

Hi, I have some trouble with an expression.
The expression i'm supposed to solve is
find the derivative of
$\displaystyle f(x)=(x^(sin(x)))*(x2+1)^(-3x)$

d/dx u(x)^(v(x)) = ??

found something on an online calculator suggesting that the formula for a chain powered by a chain is:
d/dx u(x)^(v(x)) = u(x)^(v(x))*d/dx[ln(u(x))*(v(x))]

anyone here who have solved anything like this before?

Last edited by dorkfork; October 5th, 2016 at 08:26 AM.
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October 5th, 2016, 08:42 AM   #2
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I assume you mean the function ...

$f(x) = (x\sin{x})(x^2+1)^{-3x}$

let $u = x\sin{x}$

$v = (x^2+1)^{-3x}$

start by finding the derivative of $v$ using logarithmic differentiation ...

$v = (x^2+1)^{-3x}$

$\ln{v} = -3x\ln(x^2+1)$

$\dfrac{v'}{v} = -3x \cdot \dfrac{2x}{x^2+1} - 3\ln(x^2+1)$

$\dfrac{v'}{v} = -3\bigg[x \cdot \dfrac{2x}{x^2+1} + \ln(x^2+1)\bigg]$

$v' = -3v \bigg[\dfrac{2x^2}{x^2+1} + \ln(x^2+1) \bigg]$

$v' = -3(x^2+1)^{-3x} \bigg[\dfrac{2x^2}{x^2+1} + \ln(x^2+1) \bigg]$

I know, it's a mess. Continuing with $u$ ...


$u' = x\cos{x} + \sin{x}$

now you can put it all together using the product rule ...

$\dfrac{d}{dx}(u \cdot v) = uv'+u'v$
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October 7th, 2016, 08:37 AM   #3
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seems that you mean for sinx to be an exponent also.
let's do logarithmic differentiation for the whole thing ...

$f(x) = x^{\sin{x}} \cdot (x^2+1)^{-3x}$

$\ln[f(x)] = \ln \bigg[ x^{\sin{x}} \cdot (x^2+1)^{-3x} \bigg]$

using properties of logs ...

$\ln[f(x)] = \ln \left[ x^{\sin{x}}\right] + \ln\left[ (x^2+1)^{-3x} \right]$

$\ln[f(x)] = \sin{x} \cdot \ln{x} - 3x \cdot \ln(x^2+1)$

take the derivative of both sides ...

$\dfrac{f'(x)}{f(x)} = \bigg[\dfrac{\sin{x}}{x} + \cos{x} \cdot \ln{x} \bigg] - \bigg[3x \cdot \dfrac{2x}{x^2+1} + 3 \cdot \ln(x^2+1)\bigg]$

$f'(x) = f(x) \bigg(\bigg[\dfrac{\sin{x}}{x} + \cos{x} \cdot \ln{x} \bigg] - 3\bigg[\dfrac{2x^2}{x^2+1} + \ln(x^2+1)\bigg] \bigg)$
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