Calculus of Trigonometric Functions

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Differentiating Various Trig Functions (sin/cos)

Lesson

In differential calculus, we begin the study of the subject by establishing from first principles the rules that subsequently enable us to differentiate a wide range of (possibly) complicated functions with a minimum of labour.

The rules that we can use (because they have previously been shown to be valid) include the following. In all cases, we consider smoothly varying continuous functions.

Remember!

The derivative of a sum of terms is the sum of the derivatives of the terms.

The derivative of a constant function is zero.

If $y=x^k$`y`=`x``k`, then $y'(x)=kx^{k-1}$`y`′(`x`)=`k``x``k`−1 ($k$`k` can be any real number).

If a function $h$`h` is the product of functions $f$`f` and $g$`g`, then $h'=f'g+fg'$`h`′=`f`′`g`+`f``g`′.

If $h$`h` is a quotient, $h=\frac{f}{g}$`h`=`f``g`, then $h'=\frac{f'g-fg'}{g^2}$`h`′=`f`′`g`−`f``g`′`g`2.

If $h$`h` is a composite function, $f\circ g$`f`∘`g` (a function of a function), then $h'=f'(g)g'$`h`′=`f`′(`g`)`g`′. This is also expressed as the *chain rule* using Leibniz notation: If $h(x)=f\left(g(x)\right)$`h`(`x`)=`f`(`g`(`x`)), then $\frac{dh}{dx}=\frac{dh}{dg}\frac{dg}{dx}$`d``h``d``x`=`d``h``d``g``d``g``d``x`.

If $y$`y` is the sine function, $y'$`y`′ is the cosine function.

If $y$`y` is the cosine function, $y'$`y`′ is the negative of the sine function.

Find the gradient function for $y=\sin2x\cos x$`y`=`s``i``n`2`x``c``o``s``x`.

We will need the product rule and the function of a function rule. To differentiate $\sin2x$`s``i``n`2`x` we first differentiate the 'outside function', the sine function, and then the 'inside function', the $\times2$×2 function. So, the derivative of $\sin2x$`s``i``n`2`x` is $\cos2x\times2$`c``o``s`2`x`×2, or just $2\cos2x$2`c``o``s`2`x`.

Then, using the product rule, we have

$y'$y′ |
$=$= | $2\cos2x.\cos x+\sin2x.(-1)\sin x$2cos2x.cosx+sin2x.(−1)sinx |

$=$= | $2\cos2x.\cos x-\sin2x\sin x$2cos2x.cosx−sin2xsinx |

Differentiate $\frac{\cos x}{4x}$`c``o``s``x`4`x`.

To differentiate the $4x$4`x` part of this expression we need the rule (not stated above) that says the derivative of a constant multiplied by a variable is just the constant. We can get this rule from the rules given above by thinking of the $4x$4`x` as a sum $x+x+x+x$`x`+`x`+`x`+`x` which must have derivative $1+1+1+1=4$1+1+1+1=4.

Then, using the quotient rule, we have the derivative $\frac{-\sin x\times(4x)-\cos x\times4}{\left(4x\right)^2}$−`s``i``n``x`×(4`x`)−`c``o``s``x`×4(4`x`)2, which simplifies to $\frac{-\left(x\sin x+\cos x\right)}{4x^2}$−(`x``s``i``n``x`+`c``o``s``x`)4`x`2.

Differentiate $y=\tan\left(x^\circ\right)$`y`=`t``a``n`(`x`°).

Here the variable $x$`x` is expressed in degrees. The given derivatives of the trigonometric functions assume radian measure. So, we rewrite the function as $y=\tan\frac{\pi x}{180}$`y`=`t``a``n`π`x`180.

To differentiate the tangent function, we express it as $\frac{\text{sine}}{\text{cosine}}$sinecosine, which is how the tangent function is defined.

Thus, $u=\tan\theta=\frac{\sin\theta}{\cos\theta}$`u`=`t``a``n``θ`=`s``i``n``θ``c``o``s``θ` has derivative $u'=\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}=\sec^2\theta$`u`′=`c``o``s`2`θ`+`s``i``n`2`θ``c``o``s`2`θ`=1`c``o``s`2`θ`=`s``e``c`2`θ`.

Putting this together, we have $y'=\frac{\pi}{180}\sec^2\frac{\pi x}{180}$`y`′=π180`s``e``c`2π`x`180

Expressed in degrees, this is $y'=\frac{\pi}{180}\sec^2x$`y`′=π180`s``e``c`2`x` and we see that an incorrect result would have been obtained had we not first converted from degrees to radians.

Differentiate $y=3x+\sin5x$`y`=3`x`+`s``i``n`5`x`.

Differentiate $y=\cos\left(\frac{\pi t}{2}-\frac{\pi}{3}\right)$`y`=`c``o``s`(π`t`2−π3).

Consider the function $y=\frac{\sin^2\left(x\right)}{\cos x}$`y`=`s``i``n`2(`x`)`c``o``s``x`.

Prove that $\frac{\sin^2\left(x\right)}{\cos x}=\sec x-\cos x$

`s``i``n`2(`x`)`c``o``s``x`=`s``e``c``x`−`c``o``s``x`.Differentiate $f\left(x\right)=\sec x$

`f`(`x`)=`s``e``c``x`.Hence find $\frac{dy}{dx}$

`d``y``d``x` without using the product or quotient rule.

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply differentiation methods in solving problems