Page:Calculus Made Easy.pdf/191

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 $\cos \theta$ gives us $-\sin \theta$, and differentiating $-\sin \theta$ gives us $-\cos \theta$; or thus:

${\frac {d^{2}(\cos \theta )}{d\theta ^{2}}}=-\cos \theta$.

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

 

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Examples.

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

(1) $y=\arcsin x$.

If $y$ is the arc whose sine is $x$, then $x=\sin y$.

${\frac {dx}{dy}}=\cos y$.

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

${\begin{aligned}{\frac {dy}{dx}}&={\frac {1}{\;{\dfrac {dx}{dy}}\;}}={\frac {1}{\cos y}}.\\{\text{Now}}\quad \quad \quad \cos y&={\sqrt {1-\sin ^{2}y}}={\sqrt {1-x^{2}}};\\{\text{hence}}\quad \quad \quad {\frac {dy}{dx}}&={\frac {1}{\sqrt {1-x^{2}}}},\end{aligned}}$

a rather unexpected result.