9. Introducing a Useful Dodge

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.

Example 1
Differentiate \(y = \sqrt{a+x}\).
Solution
Let \(a+x = u\). \[\begin{aligned} \frac{du}{dx} &= 1;\quad y=u^{\frac{1}{2}};\quad \frac{dy}{du} = \tfrac{1}{2} u^{-\frac{1}{2}} = \tfrac{1}{2} (a+x)^{-\frac{1}{2}}.\\  \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx} = \frac{1}{2\sqrt{a+x}}.\end{aligned}\]
Example 2
Differentiate \(y = \dfrac{1}{\sqrt{a+x^2}}\).
Solution
Let \(a + x^2 = u\). \[\begin{aligned} \frac{du}{dx} &= 2x;\quad y=u^{-\frac{1}{2}};\quad \frac{dy}{du} = -\tfrac{1}{2}u^{-\frac{3}{2}}.\\  \frac{dy}{dx} &= \frac{dy}{du}\times \frac{du}{dx} = – \frac{x}{\sqrt{(a+x^2)^3}}.\end{aligned}\]

Example 3
Differentiate \(y = \left(m – nx^{\frac{2}{3}} + \dfrac{p}{x^{\frac{4}{3}}}\right)^a\).
Solution
Let \(m – nx^{\frac{2}{3}} + px^{-\frac{4}{3}} = u\). \[\begin{gathered} \frac{du}{dx} = -\tfrac{2}{3} nx^{-\frac{1}{3}} – \tfrac{4}{3} px^{-\frac{7}{3}};\\  y = u^a;\quad \frac{dy}{du} = a u^{a-1}. \\  \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = -a\left(m -nx^{\frac{2}{3}} + \frac{p}{x^{\frac{4}{3}}}\right)^{a-1} (\tfrac{2}{3} nx^{-\frac{1}{3}} + \tfrac{4}{3} px^{-\frac{7}{3}}).\end{gathered}\]

Example 4
Differentiate \(y=\dfrac{1}{\sqrt{x^3 – a^2}}\).

Solution
Let \(u = x^3 – a^2\). \[\begin{aligned} \frac{du}{dx} &= 3x^2;\quad y = u^{-\frac{1}{2}};\quad \frac{dy}{du}=-\frac{1}{2}(x^3 – a^2)^{-\frac{3}{2}}. \\ \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx} = -\frac{3x^2}{2\sqrt{(x^3 – a^2)^3}}.\end{aligned}\]

Example 5
Differentiate \(y=\sqrt{\dfrac{1-x}{1+x}}\).

Solution
Write this as \(y=\dfrac{(1-x)^{\frac{1}{2}}}{(1+x)^{\frac{1}{2}}}\). \[\frac{dy}{dx} = \frac{(1+x)^{\frac{1}{2}}\, \dfrac{d(1-x)^{\frac{1}{2}}}{dx} – (1-x)^{\frac{1}{2}}\, \dfrac{d(1+x)^{\frac{1}{2}}}{dx}}{1+x}.\]

(We may also write \(y = (1-x)^{\frac{1}{2}} (1+x)^{-\frac{1}{2}}\) and differentiate as a product.)

Proceeding as in  (1) above, we get \[\frac{d(1-x)^{\frac{1}{2}}}{dx} = -\frac{1}{2\sqrt{1-x}}; \quad\text{and}\quad \frac{d(1+x)^{\frac{1}{2}}}{dx} = \frac{1}{2\sqrt{1+x}}.\]

Hence

\[\begin{aligned} \frac{dy}{dx} &= – \frac{(1 + x)^{\frac{1}{2}}}{2(1 + x)\sqrt{1-x}} – \frac{(1 – x)^{\frac{1}{2}}}{2(1 + x)\sqrt{1+x}} \\ &= – \frac{1}{2\sqrt{1+x}\sqrt{1-x}} – \frac{\sqrt{1-x}}{2 \sqrt{(1+x)^3}};\\ \text{or   } \frac{dy}{dx} &= – \frac{1}{(1+x)\sqrt{1-x^2}}.\end{aligned}\]

Example 6
Differentiate \(y = \sqrt{\dfrac{x^3}{1+x^2}}\).

Solution
We may write this \[\begin{gathered} y = x^{\frac{3}{2}}(1+x^2)^{-\frac{1}{2}}; \\ \frac{dy}{dx} = \tfrac{3}{2} x^{\frac{1}{2}}(1 + x^2)^{-\frac{1}{2}} + x^{\frac{3}{2}} \times \frac{d\bigl[(1+x^2)^{-\frac{1}{2}}\bigr]}{dx}.\end{gathered}\]

Differentiating \((1+x^2)^{-\frac{1}{2}}\), as shown in  (2) above, we get \[\frac{d\bigl[(1+x^2)^{-\frac{1}{2}}\bigr]}{dx} = – \frac{x}{\sqrt{(1+x^2)^3}};\] so that \[\frac{dy}{dx} = \frac{3\sqrt{x}}{2\sqrt{1+x^2}} – \frac{\sqrt{x^5}}{\sqrt{(1+x^2)^3}} = \frac{\sqrt{x}(3+x^2)}{2\sqrt{(1+x^2)^3}}.\]

Example 7
Differentiate \(y=(x+\sqrt{x^2+x+a})^3\).

Solution
Let \(x+\sqrt{x^2+x+a}=u\). \[\begin{gathered} \frac{du}{dx} = 1 + \frac{d\bigl[(x^2+x+a)^{\frac{1}{2}}\bigr]}{dx}. \\ y = u^3;\quad\text{and}\quad \frac{dy}{du} = 3u^2= 3\left(x+\sqrt{x^2+x+a}\right)^2.\end{gathered}\]

Now let \((x^2+x+a)^{\frac{1}{2}}=v\) and \((x^2+x+a) = w\).

\[ \begin{aligned} \frac{dw}{dx} &= 2x+1;\quad v = w^{\frac{1}{2}};\quad \frac{dv}{dw} = \tfrac{1}{2}w^{-\frac{1}{2}}. \\ \frac{dv}{dx} &= \frac{dv}{dw} \times \frac{dw}{dx} = \tfrac{1}{2}(x^2+x+a)^{-\frac{1}{2}}(2x+1). \\ \text{Hence  } \frac{du}{dx} &= 1 + \frac{2x+1}{2\sqrt{x^2+x+a}}, \\ \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx}\\ &= 3\left(x+\sqrt{x^2+x+a}\right)^2 \left(1 +\frac{2x+1}{2\sqrt{x^2+x+a}}\right). \end{aligned}\]

Example 8
Differentiate \(y=\sqrt{\dfrac{a^2+x^2}{a^2-x^2}} \sqrt[3]{\dfrac{a^2-x^2}{a^2+x^2}}\).

Solution
We get \[\begin{aligned} y &= \frac{(a^2+x^2)^{\frac{1}{2}} (a^2-x^2)^{\frac{1}{3}}} {(a^2-x^2)^{\frac{1}{2}} (a^2+x^2)^{\frac{1}{3}}} = (a^2+x^2)^{\frac{1}{6}} (a^2-x^2)^{-\frac{1}{6}}. \\ \frac{dy}{dx} &= (a^2+x^2)^{\frac{1}{6}} \frac{d\bigl[(a^2-x^2)^{-\frac{1}{6}}\bigr]}{dx} + \frac{d\bigl[(a^2+x^2)^{\frac{1}{6}}\bigr]}{(a^2-x^2)^{\frac{1}{6}}\, dx}.\end{aligned}\]

Let \(u = (a^2-x^2)^{-\frac{1}{6}}\) and \(v = (a^2 – x^2)\). \[\begin{aligned} u &= v^{-\frac{1}{6}};\quad \frac{du}{dv} = -\frac{1}{6}v^{-\frac{7}{6}};\quad \frac{dv}{dx} = -2x. \\  \frac{du}{dx} &= \frac{du}{dv} \times \frac{dv}{dx} = \frac{1}{3}x(a^2-x^2)^{-\frac{7}{6}}.\end{aligned}\]

Let \(w = (a^2 + x^2)^{\frac{1}{6}}\) and \(z = (a^2 + x^2)\). \[\begin{aligned} w &= z^{\frac{1}{6}};\quad \frac{dw}{dz} = \frac{1}{6}z^{-\frac{5}{6}};\quad \frac{dz}{dx} = 2x. \\  \frac{dw}{dx} &= \frac{dw}{dz} \times \frac{dz}{dx} = \frac{1}{3} x(a^2 + x^2)^{-\frac{5}{6}}.\end{aligned}\]

Hence

\[ \begin{aligned} \frac{dy}{dx} &= (a^2+x^2)^{\frac{1}{6}} \frac{x}{3(a^2-x^2)^{\frac{7}{6}}} + \frac{x}{3(a^2-x^2)^{\frac{1}{6}} (a^2+x^2)^{\frac{5}{6}}}; \\ \text{or  } \frac{dy}{dx} &= \frac{x}{3} \left[\sqrt[6]{\frac{a^2+x^2}{(a^2-x^2)^7}} + \frac{1}{\sqrt[6]{(a^2-x^2)(a^2+x^2)^5]}} \right]. \end{aligned}\]

Example 9
Differentiate \(y^n\) with respect to \(y^5\).

Solution
\[\frac{d(y^n)}{d(y^5)} = \frac{ny^{n-1}}{5y^{5-1}} = \frac{n}{5} y^{n-5}.\]

Example 10
Find the first and second differential coefficients of \(y = \dfrac{x}{b} \sqrt{(a-x)x}\).

Solution

\[\frac{dy}{dx} = \frac{x}{b}\, \frac{d\bigl\{\bigl[(a-x)x\bigr]^{\frac{1}{2}}\bigr\}}{dx} + \frac{\sqrt{(a-x)x}}{b}.\]

Let \(\bigl[(a-x)x\bigr]^{\frac{1}{2}} = u\) and let \((a-x)x = w\); then \(u = w^{\frac{1}{2}}\).

Hence \[\frac{dy}{dx} = \frac{x(a-2x)}{2b\sqrt{(a-x)x}} + \frac{\sqrt{(a-x)x}}{b} = \frac{x(3a-4x)}{2b\sqrt{(a-x)x}}.\]

Now \[\begin{aligned} \frac{d^2y}{dx^2} &= \frac{2b \sqrt{(a-x)x}\, (3a-8x) – \dfrac{(3ax-4x^2)b(a-2x)}{\sqrt{(a-x)x}}} {4b^2(a-x)x} \\ &= \frac{3a^2-12ax+8x^2}{4b(a-x)\sqrt{(a-x)x}}.\end{aligned}\]

(We shall need these two last differential coefficients later on. See chapter 12 .)

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 \(y = \dfrac{\sqrt{1 – \theta^2}}{1 – \theta}\).

Answers to Exercises
(1) \(\dfrac{x}{\sqrt{ x^2 + 1}}\). (2) \(\dfrac{x}{\sqrt{ x^2 + a^2}}\). (3) \(- \dfrac{1}{2 \sqrt{(a + x)^3}}\).
(4) \(\dfrac{ax}{\sqrt{(a – x^2)^3}}\). (5) \(\dfrac{2a^2 – x^2}{x^3 \sqrt{ x^2 – a^2}}\). (6) \(\dfrac{\frac{3}{2} x^2 \left[ \frac{8}{9} x \left( x^3 + a \right) – \left( x^4 + a \right) \right]}{(x^4 + a)^{\frac{2}{3}} (x^3 + a)^{\frac{3}{2}}}\)
(7) \(\dfrac{2a \left(x – a \right)}{(x + a)^3}\).
 (8) \(\frac{5}{2} y^3\). (9) \(\dfrac{1}{(1 – \theta) \sqrt{1 – \theta^2}}\).

 

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.

Example 1
If \(z = 3x^4\);\(v = \dfrac{7}{z^2}\);\(y =\sqrt{1+v}\), find \(\dfrac{dv}{dx}\).

Solution
We have \[\begin{aligned} \frac{dy}{dv} &= \frac{1}{2\sqrt{1+v}};\quad \frac{dv}{dz} = -\frac{14}{z^3};\quad \frac{dz}{dx} = 12x^3. \\  \frac{dy}{dx} &= -\frac{168x^3}{(2\sqrt{1+v})z^3} = -\frac{28}{3x^5\sqrt{9x^8+7}}.\end{aligned}\]

Example 2
If \(t = \dfrac{1}{5\sqrt{\theta}}\);\(x = t^3 + \dfrac{t}{2}\);\(v = \dfrac{7x^2}{\sqrt[3]{x-1}}\), find \(\dfrac{dv}{d\theta}\).

Solution

\[ \begin{aligned} \frac{dv}{dx} = \frac{7x(5x-6)}{3\sqrt[3]{(x-1)^4}};\quad \frac{dx}{dt} = 3t^2 + \tfrac{1}{2};\quad \frac{dt}{d\theta} = -\frac{1}{10\sqrt{\theta^3}}. \\ \text{Hence } \frac{dv}{d\theta} = -\frac{7x(5x-6)(3t^2+\frac{1}{2})} {30\sqrt[3]{(x-1)^4} \sqrt{\theta^3}}, \end{aligned}\]

an expression in which \(x\) must be replaced by its value, and \(t\) by its value in terms of \(\theta\).

Example 3
  If \(\theta = \dfrac{3a^2x}{\sqrt{x^3}}\);\(\omega = \dfrac{\sqrt{1-\theta^2}}{1+\theta}\);and \(\phi = \sqrt{3} – \dfrac{1}{\omega\sqrt{2}}\), find \(\dfrac{d\phi}{dx}\).

Solution
We get \[\begin{gathered} \theta = 3a^2x^{-\frac{1}{2}};\quad \omega = \sqrt{\frac{1-\theta}{1+\theta}};\quad \text{and}\quad \phi = \sqrt{3}- \frac{1}{\sqrt{2}} \omega^{-1}. \\ \frac{d\theta}{dx} = -\frac{3a^2}{2\sqrt{x^3}};\quad \frac{d\omega}{d\theta} = -\frac{1}{(1+\theta)\sqrt{1-\theta^2}}\end{gathered}\] (see example 5); and \[\frac{d\phi}{d\omega} = \frac{1}{\sqrt{2}\omega^2}.\]

So that \(\dfrac{d\theta}{dx} = \dfrac{1}{\sqrt{2} \times \omega^2} \times \dfrac{1}{(1+\theta) \sqrt{1-\theta^2}} \times \dfrac{3a^2}{2\sqrt{x^3}}\).

Replace now first \(\omega\), then \(\theta\) by its value.

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 \(\dfrac{du}{dx}\).

 

Answers to Exercises

(1) \(\dfrac{dw}{dx} = \dfrac{3x^2 \left( 3 + 3x^3 \right)} {27 \left(\frac{1}{2} x^3 + \frac{1}{4} x^6 \right)^3}\).

(2) \( \dfrac{dv}{dx} = – \dfrac{12x}{\sqrt{1 + \sqrt{2} + 3x^2} \left(\sqrt{3} + 4 \sqrt{1 + \sqrt{2} + 3x^2}\right)^2} \).

(3) \( \dfrac{du}{dx} = – \dfrac{x^2 \left(\sqrt{3} + x^3 \right)} {\sqrt{ \left[ 1 + \left( 1 + \dfrac{x^3}{\sqrt{3}} \right) ^2 \right]^3}} \).