anonymous
  • anonymous
Use the chain rule to find the derivative of the following function:
Mathematics
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[f(x) = e ^{\cos(x ^{3)}}\]
blockcolder
  • blockcolder
It's gonna get ugly, but I'll start things. For any function u(x), we have, using the chain rule: \[{d \over dx}(e^u)=e^u {du \over dx}\]. So in this case, u=cos(x^3). Now, use the chain rule again to find du/dx.
Zarkon
  • Zarkon
\[\frac{d}{dx}[f(g(h(x)))]=f'(g(h(x)))g'(h(x))h'(x)\]

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anonymous
  • anonymous
\[f(x) = e^{\cos \left( x^3 \right)}\]\[v = x^3\]\[u = \cos \left( x^3 \right)=\cos v\]\[f(x) = y = e^{u}\]\[f'(x) = \frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}\]\[f'(x) = \frac{d}{du}\left( e^u \right)\frac{d}{dv}\left( \cos v \right)\frac{d}{dx}\left( x^3 \right)\]
anonymous
  • anonymous
( e^u ) ' = u' * e^u = ( cos x³ )' * e^ ( cosx³) = -3x² sinx³ * e^ ( cosx³)
anonymous
  • anonymous
e^(cos(x^3)) Well, we shall take it one by one. If you look closely, the expression is a function within a function within a function. We can look at the outermost term to be an exponential function. We know that deriv of e^x = e^x still. But x in here is cos (x^3), so the derivative of the outermost is e^(cos(x^3)) still. Now the function inside the outermost function is cos (x^3). Since e is usually raised to just x, this function wherein x is also a function should also be differentiated. The derivative of cos (x^3) = -sin(x^3). Since we know that deriv of cos x = -sin x, where x in this case is x^3. Next, we have the third layer of functions (x^3). This is easily differentiated. It's actually 3x^2. This would have to be the end of the layers of functions since the derivative of x is just 1. We have everything and the last step is to just multiply everything. Therefore, the derivative of e^(cos(x^3)) using CR = e^(cos(x^3))(-sin(x^3))(3x^2).

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