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Calculating the derivative of a function of the form $\\tan(a \\ln(bx))$ using the chain rule.

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Calculate the derivative of $y=\\simplify[all,fractionNumbers]{tan({a}ln({b}x))}$.

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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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• Let $u=g(x)$;
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• Rewrite $y$ in terms of $u$, such that $y=f(u)$;
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• Calculate $\\frac{du}{dx}$ and $\\frac{dy}{du}$;
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• Write $\\frac{dy}{dx}$ as a product of $\\frac{du}{dx}$ and $\\frac{dy}{du}$;
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• Make sure $\\frac{dy}{dx}$ is only in terms of $x$. Ensure any $u$ terms have been replaced using the initial substitution.
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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 $y=\\simplify[all,fractionNumbers]{tan({a}ln({b}x))}$, \$g(x)=\\simplify[all, fractionNumbers, unitFactor]{{a}ln({b}x)}.\$

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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]{tan(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}/x}, \\quad \\frac{dy}{du}=\\simplify[all, fractionNumbers, unitFactor]{sec(u)^2}.\$

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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}/x} \\times\\simplify[all, fractionNumbers, unitFactor]{sec(u)^2}. \\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 $u=\\simplify[all, fractionNumbers, unitFactor]{{a}ln({b}x)}$:

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\$\\frac{dy}{dx} =\\simplify[all,fractionNumbers]{{a}/x sec({a}ln({b}x))^2}.\$

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

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