Differentiating Various Trig Functions (sin/cos)
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.
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=xk, then $y'(x)=kx^{k-1}$y′(x)=kxk−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+fg′.
If $h$h is a quotient, $h=\frac{f}{g}$h=fg, then $h'=\frac{f'g-fg'}{g^2}$h′=f′g−fg′g2.
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 $\frac{dh}{dx}=\frac{dh}{dg}\frac{dg}{dx}$dhdx=dhdgdgdx.
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.
Example 1
Find the gradient function for $y=\sin2x\cos x$y=sin2xcosx.
We will need the product rule and the function of a function rule. To differentiate $\sin2x$sin2x we first differentiate the 'outside function', the sine function, and then the 'inside function', the $\times2$×2 function. So, the derivative of $\sin2x$sin2x is $\cos2x\times2$cos2x×2, or just $2\cos2x$2cos2x.
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 |
Example 2
Differentiate $\frac{\cos x}{4x}$cosx4x.
To differentiate the $4x$4x 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$4x 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 $\frac{-\sin x\times(4x)-\cos x\times4}{\left(4x\right)^2}$−sinx×(4x)−cosx×4(4x)2, which simplifies to $\frac{-\left(x\sin x+\cos x\right)}{4x^2}$−(xsinx+cosx)4x2.
Example 3
Differentiate $y=\tan\left(x^\circ\right)$y=tan(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=tanπx180.
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=tanθ=sinθcosθ has derivative $u'=\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}=\sec^2\theta$u′=cos2θ+sin2θcos2θ=1cos2θ=sec2θ.
Putting this together, we have $y'=\frac{\pi}{180}\sec^2\frac{\pi x}{180}$y′=π180sec2πx180
Expressed in degrees, this is $y'=\frac{\pi}{180}\sec^2x$y′=π180sec2x and we see that an incorrect result would have been obtained had we not first converted from degrees to radians.
More Examples
Example 4
Differentiate $y=3x+\sin5x$y=3x+sin5x.
Example 5
Differentiate $y=\cos\left(\frac{\pi t}{2}-\frac{\pi}{3}\right)$y=cos(πt2−π3).
Example 6
Consider the function $y=\frac{\sin^2\left(x\right)}{\cos x}$y=sin2(x)cosx.
Prove that $\frac{\sin^2\left(x\right)}{\cos x}=\sec x-\cos x$sin2(x)cosx=secx−cosx.
Differentiate $f\left(x\right)=\sec x$f(x)=secx.
Hence find $\frac{dy}{dx}$dydx without using the product or quotient rule.