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coefficients

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Differentiate the following trigonometric functions using the chain rule.

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Do not write out $dy/dx$; only input the differentiated right hand side of each equation.

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$y=\\sin(\\var{c[0]}x)$

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$\\frac{dy}{dx}=$ [[0]]

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$y=\\cos(\\var{c[1]}x)$

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$\\frac{dy}{dx}=$ [[0]]

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$y=-\\sin(\\var{c[2]}x^2)$

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$\\frac{dy}{dx}=$ [[0]]

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$y=-5\\cos(\\var{c[3]}x)+\\sin(\\var{c[4]}x)$

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$\\frac{dy}{dx}=$ [[0]]

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$y=\\tan^\\var{p[0]}(x)$

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$\\frac{dy}{dx}=$ [[0]]

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$y=\\cos(x^\\var{p[1]}-1)$

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$\\frac{dy}{dx}=$ [[0]]

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If you don't know how to differentiate trigonometric functions, please see 'Differentiation 4 - Trigonometric Functions'.

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These questions use the chain rule.

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The earlier questions are easy to do by inspection, e.g using Part a:

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$y=sin(\\var{c[0]}x)$.

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We differentiate the term(s) inside the function, here the term is $\\var{c[0]}x$.

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Then we derive $sin$ of any function, giving us $cos$.

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Putting our results together, we get

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$\\var{c[0]}cos(\\var{c[0]}x)$.

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We will now go through an entire worked example of the formal method of the chain rule using Part e.

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The expression we will be differentiating here is

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$y=tan^\\var{p[0]}(x)$.

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As a reminder, the chain rule is defined as

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$\\frac{dy}{dx}=\\frac{dy}{du}\\times\\frac{du}{dx}$.

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Now we let $u=tanx$, so then $y=u^\\var{p[0]}$

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This becomes an easy differentiation using $\\frac{dy}{du}\\times\\frac{du}{dx}$:

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Differentiate $y$ with respect to $u$, giving $\\simplify{{p[0]}u^{{p[0]}-1}}$.

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Then differentiate $u$ with respect to $x$, giving $sec^2x$.

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Multiply these results together, and substitue $tan$ back in for $u$.

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$\\simplify{{p[0]}(tan^{{p[0]}-1}(x))*sec^2(x)}$.