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\nDo not write out $dy/dx$; only input the differentiated right hand side of each equation.
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\n$\\frac{dy}{dx}=$ [[0]]
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\n$\\frac{dy}{dx}=$ [[0]]
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\n$\\frac{dy}{dx}=$ [[0]]
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\n$\\frac{dy}{dx}=$ [[0]]
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\n$\\frac{dy}{dx}=$ [[0]]
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\n$\\frac{dy}{dx}=$ [[0]]
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\n\n
These questions use the chain rule.
\nThe earlier questions are easy to do by inspection, e.g using Part a:
\n$y=sin(\\var{c[0]}x)$.
\nWe differentiate the term(s) inside the function, here the term is $\\var{c[0]}x$.
\nThen we derive $sin$ of any function, giving us $cos$.
\nPutting our results together, we get
\n$\\var{c[0]}cos(\\var{c[0]}x)$.
\n\n\n\nWe will now go through an entire worked example of the formal method of the chain rule using Part e.
\nThe expression we will be differentiating here is
\n$y=tan^\\var{p[0]}(x)$.
\nAs a reminder, the chain rule is defined as
\n$\\frac{dy}{dx}=\\frac{dy}{du}\\times\\frac{du}{dx}$.
\nNow we let $u=tanx$, so then $y=u^\\var{p[0]}$
\nThis becomes an easy differentiation using $\\frac{dy}{du}\\times\\frac{du}{dx}$:
\nDifferentiate $y$ with respect to $u$, giving $\\simplify{{p[0]}u^{{p[0]}-1}}$.
\nThen differentiate $u$ with respect to $x$, giving $sec^2x$.
\nMultiply these results together, and substitue $tan$ back in for $u$.
\nYour final result is therefore
\n$\\simplify{{p[0]}(tan^{{p[0]}-1}(x))*sec^2(x)}$.
", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "More work on differentiation with trigonometric functions
"}, "contributors": [{"name": "Katie Lester", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/586/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}]}], "contributors": [{"name": "Katie Lester", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/586/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}