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Calculating the derivative of a function of the form $e^{\\cos(ax+b)}$ using the chain rule.

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Calculate the derivative of $y=\\simplify[all,fractionNumbers]{e^(cos({a}*x+{b}))}$.

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If we have a function of the form $y=f(g(x))$, sometimes described as a function of a function, to calculate its derivative we need to use the chain rule:

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\\[ \\frac{dy}{dx} = \\frac{du}{dx} \\times \\frac{dy}{du}.\\]

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This can be split up into steps:

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Following this process, we must first identify $g(x)$. Since the function is of the form $y=f(g(x))$, we are looking for the 'inner' function.

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So, for $y=\\simplify[all,fractionNumbers]{e^(cos({a}*x+{b}))}$, \\[g(x)=\\simplify[all, fractionNumbers, unitFactor]{cos({a}*x+{b})}.\\]

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If we now set $u=g(x)$, we can rewrite $y$ in terms of $u$ such that $y=f(u)$:

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\\[y=\\simplify[all, fractionNumbers,unitFactor]{e^(u)}.\\]

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Next, we calculate the two derivatives $\\frac{du}{dx}$ and $\\frac{dy}{du}$:

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\\[\\frac{du}{dx}=\\simplify[all,fractionNumbers]{-{a}sin({a}x+{b})}, \\quad \\frac{dy}{du}=\\simplify[all, fractionNumbers, unitFactor]{e^u}.\\]

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Plugging these into the chain rule:

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\\[ \\begin{split} \\frac{dy}{dx} &= \\frac{du}{dx} \\times \\frac{dy}{du}, \\\\&=\\simplify[all,fractionNumbers]{-{a}sin({a}x+{b})}\\times\\simplify[all, fractionNumbers, unitFactor]{e^u}. \\end{split} \\]

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Finally, we need to express $\\frac{dy}{dx}$ only in terms of $x$, so we must replace the $u$ term using the initial substitution $u=\\simplify[all, fractionNumbers, unitFactor]{cos({a}x+{b})}$:

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\\[ \\frac{dy}{dx} =\\simplify[all,fractionNumbers]{-{a}sin({a}x+{b})e^(cos({a}x+{b}))}.\\]

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$\\frac{dy}{dx}=$[[0]]

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