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For any questions on differentiating trigonometric functions, study the following table of derivatives provided by mathcentre: http://www.mathcentre.ac.uk/resources/Engineering%20maths%20first%20aid%20kit/latexsource%20and%20diagrams/8_2.pdf

These questions use the chain rule.

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

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

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

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

Putting our results together, we get

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

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

The expression we will be differentiating here is

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

As a reminder, the chain rule is defined as

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

Now we let $u=\\tan(x)$, so then $y=u^\\var{p[0]}$

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

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

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

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

Your final result is therefore

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

Note that you have to input powers of these trig functions in the form tan(x)^m and sec(x)^n for suitable values of m and n.

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

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

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

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

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

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

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

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

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Differentiate the following trigonometric functions using the chain rule.

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

Note that if you input an expression such as $x\\sin(x)$ then you need to include a multiplication sign as in x*sin(x).

Also to input a power of a trig function such as $\\tan^5(x)$ then you input (tan(x))^5.

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coefficients

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More work on differentiation with trigonometric functions

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