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{align*} \text{for}\; \frac{dy}{dx} &= \frac{dy}{du} × \frac{du}{dx}; \\ \text{that is,}\; \frac{dy}{dx} &= \frac{3}{2} u^{\frac{1}{2}} × 2x \\ &= \tfrac{3}{2} (x^2 + a^2)^{\frac{1}{2}} × 2x \\ &= 3x(x^2 + a^2)^{\frac{1}{2}}; \end{align*} and so the trick is done.
By and by, 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.
(1) Differentiate $y = \sqrt{a+x}$.
Let $a+x = u$. \begin{align*} \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} × \frac{du}{dx} = \frac{1}{2\sqrt{a+x}}. \end{align*}
(2) Differentiate $y = \dfrac{1}{\sqrt{a+x^2}}$.
Let $a + x^2 = u$. \begin{align*} \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}×\frac{du}{dx} = - \frac{x}{\sqrt{(a+x^2)^3}}. \end{align*}
(3) Differentiate $y = \left(m - nx^{\frac{2}{3}} + \dfrac{p}{x^{\frac{4}{3}}}\right)^a$.
Let $m - nx^{\frac{2}{3}} + px^{-\frac{4}{3}} = u$. \begin{gather*} \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}×\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{gather*}
(4) Differentiate $y=\dfrac{1}{\sqrt{x^3 - a^2}}$.
Let $u = x^3 - a^2$. \begin{align*} \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} × \frac{du}{dx} = -\frac{3x^2}{2\sqrt{(x^3 - a^2)^3}}. \end{align*}
(5) Differentiate $y=\sqrt{\dfrac{1-x}{1+x}}$.
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 example (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{align*} \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}};\\ or \frac{dy}{dx} &= - \frac{1}{(1+x)\sqrt{1-x^2}}. \end{align*}
(6) Differentiate $y = \sqrt{\dfrac{x^3}{1+x^2}}$.
We may write this \begin{gather*} 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}} × \frac{d\bigl[(1+x^2)^{-\frac{1}{2}}\bigr]}{dx}. \end{gather*}
Differentiating $(1+x^2)^{-\frac{1}{2}}$, as shown in example (2) above, we get \[ \frac{d\bigl[(1+x^2)^{-\frac{1}{2}}\bigr]}{dx} = - \frac{x}{\sqrt{(1+x^2)^3}}; \] so that \[ \frac{dy}{dx} = \frac{3\sqrt{x}}{2\sqrt{1+x^2}} - \frac{\sqrt{x^5}}{\sqrt{(1+x^2)^3}} = \frac{\sqrt{x}(3+x^2)}{2\sqrt{(1+x^2)^3}}. \]
(7) Differentiate $y=(x+\sqrt{x^2+x+a})^3$.
Let $x+\sqrt{x^2+x+a}=u$. \begin{gather*} \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{gather*}
Now let $(x^2+x+a)^{\frac{1}{2}}=v$ and $(x^2+x+a) = w$. \begin{align*} \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} × \frac{dw}{dx} = \tfrac{1}{2}(x^2+x+a)^{-\frac{1}{2}}(2x+1). \\ Hence \frac{du}{dx} &= 1 + \frac{2x+1}{2\sqrt{x^2+x+a}}, \\ \frac{dy}{dx} &= \frac{dy}{du} × \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{align*}
(8) Differentiate $y=\sqrt{\dfrac{a^2+x^2}{a^2-x^2}} \sqrt[3]{\dfrac{a^2-x^2}{a^2+x^2}}$.
We get \begin{align*} 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{align*}
Let $u = (a^2-x^2)^{-\frac{1}{6}}$ and $v = (a^2 - x^2)$. \begin{align*} 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} × \frac{dv}{dx} = \frac{1}{3}x(a^2-x^2)^{-\frac{7}{6}}. \end{align*}
Let $w = (a^2 + x^2)^{\frac{1}{6}}$ and $z = (a^2 + x^2)$. \begin{align*} 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} × \frac{dz}{dx} = \frac{1}{3} x(a^2 + x^2)^{-\frac{5}{6}}. \end{align*}
Hence \begin{align*} \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}}}; \\ 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{align*}
(9) Differentiate $y^n$ with respect to $y^5$. \[ \frac{d(y^n)}{d(y^5)} = \frac{ny^{n-1}}{5y^{5-1}} = \frac{n}{5} y^{n-5}. \]
(10) Find the first and second differential coefficients of $y = \dfrac{x}{b} \sqrt{(a-x)x}$. \[ \frac{dy}{dx} = \frac{x}{b}\, \frac{d\bigl\{\bigl[(a-x)x\bigr]^{\frac{1}{2}}\bigr\}}{dx} + \frac{\sqrt{(a-x)x}}{b}. \]
Let $\bigl[(a-x)x\bigr]^{\frac{1}{2}} = u$ and let $(a-x)x = w$; then $u = w^{\frac{1}{2}}$. \[ \frac{du}{dw} = \frac{1}{2} w^{-\frac{1}{2}} = \frac{1}{2w^{\frac{1}{2}}} = \frac{1}{2\sqrt{(a-x)x}}. \] \begin{align*} &\frac{dw}{dx} = a-2x.\\ &\frac{du}{dw} × \frac{dw}{dx} = \frac{du}{dx} = \frac{a-2x}{2\sqrt{(a-x)x}}. \end{align*}
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{align*} \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{align*}
(We shall need these two last differential coefficients later on. See Chapter XII, Exercise 11)
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 $\dfrac{dy}{dx} = \dfrac{dy}{dz} × \dfrac{dz}{dv} × \dfrac{dv}{dx}$.
Examples (1) If $z = 3x^4$; $v = \dfrac{7}{z^2}$; $y =\sqrt{1+v}$, find $\dfrac{dv}{dx}$.
We have \begin{align*} \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{align*}
(2) If $t = \dfrac{1}{5\sqrt{\theta}}$; $x = t^3 + \dfrac{t}{2}$; $v = \dfrac{7x^2}{\sqrt[3]{x-1}}$, find $\dfrac{dv}{d\theta}$. \[ \frac{dv}{dx} = \frac{7x(5x-6)}{3\sqrt[3]{(x-1)^4}};\quad \frac{dx}{dt} = 3t^2 + \tfrac{1}{2};\quad \frac{dt}{d\theta} = -\frac{1}{10\sqrt{\theta^3}}. \\ Hence \; \frac{dv}{d\theta} = -\frac{7x(5x-6)(3t^2+\frac{1}{2})} {30\sqrt[3]{(x-1)^4} \sqrt{\theta^3}}, \] an expression in which $x$ must be replaced by its value, and $t$ by its value in terms of $\theta$.
(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 $\dfrac{d\phi}{dx}$.
We get \begin{gather*} \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{gather*} (see example 5, here); and \[ \frac{d\phi}{d\omega} = \frac{1}{\sqrt{2}\omega^2}. \]
So that $\dfrac{d\theta}{dx} = \dfrac{1}{\sqrt{2} × \omega^2} × \dfrac{1}{(1+\theta) \sqrt{1-\theta^2}} × \dfrac{3a^2}{2\sqrt{x^3}}$.
Replace now first $\omega$, then $\theta$ by its value.
(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}} $