Greek letters being usual to denote angles, we will take as the usual letter for any variable angle the letter θ (“theta”).

Let us consider the function y=sinθ.

What we have to investigate is the value of d(sinθ)dθ; or, in other words, if the angle θ varies, we have to find the relation between the increment of the sine and the increment of the angle, both increments being indefinitely small in themselves. Examine Fig. 43, wherein, if the radius of the circle is unity, the height of y is the sine, and θ is the angle. Now, if θ is supposed to increase by the addition to it of the small angle dθ—an element of angle—the height of y, the sine, will be increased by a small element dy. The new height y+dy will be the sine of the new angle θ+dθ, or, stating it as an equation, y+dy=sin(θ+dθ); and subtracting from this the first equation gives dy=sin(θ+dθ)sinθ.

 

The quantity on the right-hand side is the difference between two sines, and books on trigonometry tell us how to work this out. For they tell us that if M and N are two different angles, sinMsinN=2cosM+N2sinMN2.

If, then, we put M=θ+dθ for one angle, and N=θ for the other, we may write

dy=2cosθ+dθ+θ2sinθ+dθθ2,dy=2cos(θ+12dθ)sin12dθ.

But if we regard dθ as indefinitely small, then in the limit we may neglect 12dθ by comparison with θ, and may also take sin12dθ as being the same as 12dθ. The equation then becomes:

dy=2cosθ×12dθ;dy=cosθdθ,dydθ=cosθ.

The accompanying curves, Figs. 44 and 45, show, plotted to scale, the values of y=sinθ, and dydθ=cosθ, for the corresponding values of θ.

 

 


Take next the cosine.

Let y=cosθ.

Now cosθ=sin(π2θ).

Therefore dy=d(sin(π2θ))=cos(π2θ)×d(θ),=cos(π2θ)×(dθ),dydθ=cos(π2θ).

And it follows that dydθ=sinθ.

Lastly, take the tangent.

Let y=tanθ,dy=tan(θ+dθ)tanθ.

Expanding, as shown in books on trigonometry,

tan(θ+dθ)=tanθ+tandθ1tanθtandθ;dy=tanθ+tandθ1tanθtandθtanθ=(1+tan2θ)tandθ1tanθtandθ.

Now remember that if dθ is indefinitely diminished, the value of tandθ becomes identical with dθ, and tanθdθ is negligibly small compared with 1, so that the expression reduces to

dy=(1+tan2θ)dθ1,dydθ=1+tan2θ,dydθ=sec2θ.

Collecting these results, we have: ydydθsinθcosθcosθsinθtanθsec2θ

Sometimes, in mechanical and physical questions, as, for example, in simple harmonic motion and in wave-motions, we have to deal with angles that increase in proportion to the time. Thus, if T be the time of one complete period, or movement round the circle, then, since the angle all round the circle is 2π radians, or 360, the amount of angle moved through in time t, will be

θ=2πtT,in radians,θ=360tT,in degrees.

If the frequency, or number of periods per second, be denoted by n, then n=1T, and we may then write: θ=2πnt. Then we shall have y=sin2πnt.

If, now, we wish to know how the sine varies with respect to time, we must differentiate with respect, not to θ, but to t. For this we must resort to the artifice explained in Chapter 9, and put dydt=dydθdθdt.

Now dθdt will obviously be 2πn; so that dydt=cosθ×2πn=2πncos2πnt. Similarly, it follows that d(cos2πnt)dt=2πnsin2πnt.

Second Differential Coefficient of Sine or Cosine

We have seen that when sinθ is differentiated with respect to θ it becomes cosθ; and that when cosθ is differentiated with respect to θ it becomes sinθ; or, in symbols, d2(sinθ)dθ2=sinθ.

So we have this curious result that we have found a function such that if we differentiate it twice over, we get the same thing from which we started, but with the sign changed from + to .

The same thing is true for the cosine; for differentiating cosθ gives us sinθ, and differentiating sinθ gives us cosθ; or thus: d2(cosθ)dθ2=cosθ.

Sines and cosines are the only functions of which the second differential coefficient is equal (and of opposite sign to) the original function.


Examples

With what we have so far learned we can now differentiate expressions of a more complex nature.

Example 1
y=arcsinx.

If y is the arc whose sine is x, then x=siny. dxdy=cosy.

Passing now from the inverse function to the original one, we get

dydx=1dxdy=1cosy.cosy=1sin2y=1x2;dydx=11x2,

a rather unexpected result.

Example 2
y=cos3θ.

This is the same thing as y=(cosθ)3.

 Let cosθ=v;then y=v3;dydv=3v2. dvdθ=sinθ.dydθ=dydv×dvdθ=3cos2θsinθ.

Example 3
y=sin(x+a).

Let x+a=v;then y=sinv. dydv=cosv;dvdx=1anddydx=cos(x+a).

Example 4
y=lnsinθ.

Let sinθ=v;y=lnv. dydv=1v;dvdθ=cosθ;dydθ=1sinθ×cosθ=cotθ.

Example 5
y=cotθ=cosθsinθ.

dydθ=sin2θcos2θsin2θ=(1+cot2θ)=csc2θ.

Example 6
y=tan3θ.

Let 3θ=v;y=tanv;dydv=sec2v. dvdθ=3;dydθ=3sec23θ.

Example 7
y=1+3tan2θ

y=(1+3tan2θ)12.

Let 3tan2θ=v.

y=(1+v)12;dydv=121+vdvdθ=6tanθsec2θ(for, if tanθ=u,v=3u2;dvdu=6u;dudθ=sec2θ;dvdθ=6(tanθsec2θ)dydθ=6tanθsec2θ21+3tan2θ.

Example 8
y=sinxcosx.

dydx=sinx(sinx)+cosx×cosx=cos2xsin2x.


Exercises XIV

(1) Differentiate the following: (i)y=Asin(θπ2).(ii)y=sin2θ;and y=sin2θ.(iii)y=sin3θ;and y=sin3θ.

(2) Find the value of θ for which sinθ×cosθ is a maximum.

(3) Differentiate y=12πcos2πnt.

(4) If y=sinax, find dydx.

(5) Differentiate y=lncosx.

(6) Differentiate y=18.2sin(x+26).

(7) Plot the curve y=100sin(θ15); and show that the slope of the curve at θ=75 is half the maximum slope.

(8) If y=sinθsin2θ, find dydθ.

(9) If y=atanm(θn), find the differential coefficient of y with respect to θ.

(10) Differentiate y=exsin2x.

(11) Differentiate the three equations of Exercises XIII., No. 4, and compare their differential coefficients, as to whether they are equal, or nearly equal, for very small values of x, or for very large values of x, or for values of x in the neighbourhood of x=30.

(12) Differentiate the following: (i)y=secx.(ii)y=arccosx.(iii)y=arctanx.(iv)y=arcsec x.(v)y=tanx×3secx.

(13) Differentiate y=sin(2θ+3)2.3.

(14) Differentiate y=θ3+3sin(θ+3)3sinθ3θ.

(15) Find the maximum or minimum of y=θcosθ.

Answers to Exercises

(1) (i) dydθ=Acos(θπ2);

(ii) dydθ=2sinθcosθ=sin2θ and dydθ=2cos2θ;

(iii) dydθ=3sin2θcosθ and dydθ=3cos3θ.

(2) θ=45 or π4 radians.

(3) dydt=nsin2πnt.

(4) axlnacosax.

(5) cosxsinx=\cotanx

(6) 18.2cos(x+26).

(7) The slope is dydθ=100cos(θ15), which is a maximum when (θ15)=0, or θ=15; the value of the slope being then =100. When θ=75 the slope is 100cos(7515)=100cos60=100×12=50.

(8) cosθsin2θ+2cos2θsinθ=2sinθ(cos2θ+cos2θ)=2sinθ(3cos2θ1).

(9) amnθn1tanm1(θn)sec2θn.

(10) ex(sin2x+sin2x);ex(sin2x+2sin2x+2cos2x).

(11) (i)dydx=ab(x+b)2; (ii) abexb; (iii) 190×ab(b2+x2).

(12) (i) dydx=secxtanx;

(ii) dydx=11x2;

(iii) dydx=11+x2;

(iv) dydx=1xx21;

(v) dydx=3secx(3sec2x1)2.

(13) dydθ=4.6(2θ+3)1.3cos(2θ+3)2.3.

(14) dydθ=3θ2+3cos(θ+3)ln3(cosθ×3sinθ+3θ).

(15) θ=cotθ;θ=±0.86; is max. for +θ, min. for θ.