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

 (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 $$\dfrac{du}{dx}$$.

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