Differentiating Various Trig Functions (mixed functions)

When trigonometric functions are combined with functions of other kinds, by addition, by multiplication or by composition, the resulting functions will again be differentiable functions if the separate components are differentiable.

The procedures for finding the derivatives of such functions may include using the product rule, the quotient rule or the function-of-a-function rule. The theoretical basis for these rules is briefly explained in the chapter Differentiating Various Trig Functions. 

Example 1

Investigate the function $\sqrt{x}\sin x$√xsinx. 

This function is the product of two simpler functions. If $x$x belongs to the set of real numbers, then the domain of the function we are investigating must be restricted to the positive real numbers and zero. This is because $\sqrt{x}$√x is not defined on the negative real numbers.

The derivative should exist everywhere in the open interval $\left(0,\ \infty\right)$(0, ∞). 

If $y=\sqrt{x}\sin x$y=√xsinx, then, using the product rule,  $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x}\ \times\sin x+\sqrt{x}\ \times\frac{\mathrm{d}}{\mathrm{d}x}\sin x$dydx​=ddx​√x ×sinx+√x ×ddx​sinx.

This is, $y'=\frac{\sin x}{2\sqrt{x}}+\sqrt{x}\cos x$y′=sinx2√x​+√xcosx. 

To find the points where the gradient is zero, we set $\frac{\sin x}{2\sqrt{x}}+\sqrt{x}\cos x=0$sinx2√x​+√xcosx=0. That is, $\tan x=-2x$tanx=−2x.  

This cannot be solved by algebraic manipulations but we can use a numerical method to get an approximation to any desired accuracy.

Note that since $x$x is always positive, $\tan x$tanx must be negative and therefore the smallest solution is between $\frac{\pi}{2}$π2​ and $\pi$π.  Then, because $\tan x$tanx is periodic with period $\pi$π, there must exist solutions $x=\alpha+n\pi$x=α+nπ where $\alpha$α is the smallest solution and $n$n is any positive integer.

When $x=0$x=0, the function $\sqrt{x}\sin x$√xsinx is zero, but it is positive for small values of $x>0$x>0 and must therefore be increasing in this part of its domain. At $x=\pi$x=π, the function has value $0$0. Therefore, there must have been a maximum turning point between $x=0$x=0 and $x=\pi$x=π. It follows that the value $x=\alpha$x=α must correspond to a local maximum. 

The function decreases after $x=\pi$x=π but is again $0$0 at $x=2\pi$x=2π. So, there must be a minimum turning point between $x=\pi$x=π and $x=2\pi$x=2π, corresponding to the  value $x=\alpha+\pi$x=α+π. It appears that maximum and minimum turning points alternate at the points $x=\alpha+n\pi$x=α+nπ, with maxima at even values of $n$n and minima at odd values. The evidence for this is strengthened with the help of graphing software.

To find the smallest solution $x=\alpha$x=α, we look for solutions to $\tan x=-2x$tanx=−2x, or equivalently, the smallest positive member of the solution set, $x=\arctan(-2x)+n\pi$x=arctan(−2x)+nπ. That is, we need $n=1$n=1 so that $x=\arctan(-2x)+\pi$x=arctan(−2x)+π. 

Try the following iterative procedure, which can be carried out on a hand-held calculator.

Guess an initial value for $x$x, somewhere between  $\frac{\pi}{2}$π2​ and $\pi$π. If convenient, put this value in an 'alpha' memory of your calculator, $X$X, say.

Calculate $\arctan(-2X)+\pi$arctan(−2X)+π and send the result back to the alpha memory labelled $X$X. 

Iterate this calculation several times, each time using the latest $X$X-value.

The value in memory $X$X will converge rapidly to $\alpha\approx1.8366$α≈1.8366. 

When iteration produces no further change in $X$X, we can be sure that this value, $x=X$x=X, satisfies $x=\arctan(-2x)+\pi$x=arctan(−2x)+π and hence, satisfies $\tan x=-2x$tanx=−2x.

 

More Examples

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