Sometimes one is stumped by finding that the expression to be differentiated is too complicated to tackle directly.
Thus, the equation \[y = (x^2+a^2)^{\frac{3}{2}}\] is awkward to a beginner.
Now the dodge to turn the difficulty is this: Write some symbol, such as \(u\), for the expression \(x^2 + a^2\); then the equation becomes \[y = u^{\frac{3}{2}},\] which you can easily manage; for \[\frac{dy}{du} = \frac{3}{2} u^{\frac{1}{2}}.\] Then tackle the expression \[u = x^2 + a^2,\] and differentiate it with respect to \(x\), \[\frac{du}{dx} = 2x.\] Then all that remains is plain sailing;
\[\begin{aligned} \dfrac{dy}{dx} &= \dfrac{dy}{du} \times \dfrac{du}{dx}; \\ \dfrac{dy}{dx} &= \frac{3}{2} u^{\frac{1}{2}} \times 2x \\ &= \frac{3}{2} (x^2 + a^2)^{\frac{1}{2}} \times 2x \\ &= 3x(x^2 + a^2)^{\frac{1}{2}} \end{aligned} \]
and so the trick is done.
By and bye, when you have learned how to deal with sines, and cosines, and exponentials, you will find this dodge of increasing usefulness.
Examples.
Let us practise this dodge on a few examples.
Exercises VI
Differentiate the following:
(1) \(y = \sqrt{x^2 + 1}\). | (2) \(y = \sqrt{x^2+a^2}\). |
(3) \(y = \dfrac{1}{\sqrt{a+x}}\). | (4) \(y = \dfrac{a}{\sqrt{a-x^2}}\). |
(5) \(y = \dfrac{\sqrt{x^2-a^2}}{x^2}\). | (6) \(y = \dfrac{\sqrt[3]{x^4+a}}{\sqrt[2]{x^3+a}}\). |
(7) \(y = \dfrac{a^2+x^2}{(a+x)^2}\).
(8) Differentiate \(y^5\) with respect to \(y^2\).
(9) Differentiate \(y = \dfrac{\sqrt{1 – \theta^2}}{1 – \theta}\).
The process can be extended to three or more differential coefficients, so that \(\dfrac{dy}{dx} = \dfrac{dy}{dz} \times \dfrac{dz}{dv} \times \dfrac{dv}{dx}\).
Examples.
Exercises VII.
You can now successfully try the following.
(1) If \(u = \frac{1}{2}x^3\);\(v = 3(u+u^2)\);and \(w = \dfrac{1}{v^2}\), find \(\dfrac{dw}{dx}\).
(2) If \(y = 3x^2 + \sqrt{2}\);\(z = \sqrt{1+y}\);and \(v = \dfrac{1}{\sqrt{3}+4z}\), find \(\dfrac{dv}{dx}\).
(3) If \(y = \dfrac{x^3}{\sqrt{3}}\);\(z = (1+y)^2\);and \(u = \dfrac{1}{\sqrt{1+z}}\), find \(\dfrac{du}{dx}\).