6.3 Rules for differentiation | Differential calculus

6.2 Differentiation from first principles

6.3 Rules for differentiation (EMCH7)

Determining the derivative of a function from first principles requires a long calculation and it is easy to make mistakes. However, we can use this method of finding the derivative from first principles to obtain rules which make finding the derivative of a function much simpler.

Rules for differentiation

Differentiate the following from first principles:
1. \(f\left(x\right)=x\)
2. \(f\left(x\right)=-4x\)
3. \(f\left(x\right)={x}^{2}\)
4. \(f\left(x\right)=3{x}^{2}\)
5. \(f\left(x\right)=-{x}^{3}\)
6. \(f\left(x\right)=2{x}^{3}\)
7. \(f\left(x\right)=\frac{1}{x}\)
8. \(f\left(x\right)=-\frac{2}{x}\)

Complete the table:

\(f\left(x\right)\)	\({f}'\left(x\right)\)
\(x\)
\(-4x\)
\(x^{2}\)
\(3x^{2}\)
\(-x^{3}\)
\(2x^{3}\)
\(\frac{1}{x}\)
\(-\frac{2}{x}\)

Can you identify a pattern for determining the derivative?

Rules for differentiation

General rule for differentiation:
\[\frac{d}{dx}\left[{x}^{n}\right]=n{x}^{n-1}, \text{ where } n \in \mathbb{R} \text{ and } n \ne 0.\]
The derivative of a constant is equal to zero.
\[\frac{d}{dx}\left[k\right]= 0\]
The derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function.
\[\frac{d}{dx}\left[k \cdot f\left(x\right) \right]=k \frac{d}{dx}\left[ f\left(x\right) \right]\]
The derivative of a sum is equal to the sum of the derivatives.
\[\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]=\frac{d}{dx}\left[f\left(x\right) \right] + \frac{d}{dx}\left[g\left(x\right)\right]\]
The derivative of a difference is equal to the difference of the derivatives.
\[\frac{d}{dx}\left[f\left(x\right) - g\left(x\right)\right]=\frac{d}{dx}\left[f\left(x\right) \right] - \frac{d}{dx}\left[g\left(x\right)\right]\]

Video: 28WD

Worked example 11: Rules of differentiation

Use the rules of differentiation to find the derivative of each of the following:

\(y = 3x^{5}\)
\(p = \frac{1}{4}q^{2}\)
\(f(x) = 60\)
\(y = 12x^{3} + 7x\)
\(m = \frac{3}{2}n^{4} - 1\)

Apply the appropriate rules to determine the derivative

\(\frac{dy}{dx} = 3 \left(5x^{4}\right) = 15x^{4}\)
\(\frac{dp}{dq} = \frac{1}{4}\left(2q\right) = \frac{1}{2}q\)
\(f'(x) = 0\)
\(\frac{dy}{dx} = 12\left(3x^{2}\right) + 7 = 36x^{2} + 7\)
\(\frac{dm}{dn} = \frac{3}{2}\left(4n^{3}\right) - 0 = 6n^{3}\)

Worked example 12: Rules of differentiation

Differentiate the following with respect to \(t\):

\(g(t) = 4\left( t + 1 \right)^{2} \left( t -3 \right)\)
\(k(t) = \frac{(t + 2)^{3}}{\sqrt{t}}\)

Expand the expression and apply the rules of differentiation

We have not learnt a rule for differentiating a product, therefore we must expand the brackets and simplify before we can determine the derivative:

\begin{align*} g(t) &= 4\left( t + 1 \right)^{2} \left( t - 3 \right) \\ &= 4\left( t^{2} + 2t + 1 \right) \left( t - 3 \right) \\ &= 4\left( t^{3} + 2t^{2} + t - 3t^{2} - 6t - 3 \right) \\ &= 4\left( t^{3} - t^{2} - 5t - 3 \right) \\ &= 4t^{3} - 4t^{2} - 20t - 12 \\ & \\ \therefore g'(t) &= 4 \left( 3t^{2} \right) - 4\left( 2t \right) - 20 - 0 \\ &= 12t^{2} - 8t - 20 \end{align*}

Expand the expression and apply the rules of differentiation

We have not learnt a rule for differentiating a quotient, therefore we must first simplify the expression and then we can differentiate:

\begin{align*} k(t) &= \frac{(t + 2)^{3}}{\sqrt{t}} \\ &= \frac{(t + 2)(t^{2} + 4t + 4)}{\sqrt{t}} \\ &= \frac{ t^{3} + 6t^{2} + 12t + 8}{t^{\frac{1}{2}}} \\ &= t^{-\frac{1}{2}} \left( t^{3} + 6t^{2} + 12t + 8 \right) \\ &= t^{\frac{5}{2}} + 6t^{\frac{3}{2}} + 12t^{\frac{1}{2}} + 8t^{-\frac{1}{2}} \\ & \\ \therefore g'(t) &= \frac{5}{2}t^{\frac{3}{2}} + 6 \left( \frac{3}{2}t^{\frac{1}{2}} \right) + 12 \left( \frac{1}{2} t^{-\frac{1}{2}} \right) + 8 \left( - \frac{1}{2} t^{-\frac{3}{2}} \right) \\ &= \frac{5}{2}t^{\frac{3}{2}} + 9t^{\frac{1}{2}} + 6t^{-\frac{1}{2}} - 4t^{-\frac{3}{2}} \end{align*}

Important: always write the final answer with positive exponents.

\begin{align*} g'(t) &= \frac{5}{2}t^{\frac{3}{2}} + 9t^{\frac{1}{2}} + \frac{6}{t^{\frac{1}{2}}} - \frac{4}{t^{\frac{3}{2}}} \end{align*}

When to use the rules for differentiation:

If the question does not specify how we must determine the derivative, then we use the rules for differentiation.

When to differentiate using first principles:

If the question specifically states to use first principles.
If we are required to differentiate using the definition of a derivative, then we use first principles.

Rules for differentiation

Textbook Exercise 6.4

\(y=3x^{2}\)

\(\frac{dy}{dx}=6x\)

\(f(x)=25x\)

\(f'(x)=\text{25}\)

\(k(x)=-30\)

\(k'(x)=\text{0}\)

\(y=-4x^{5}+2\)

\(\frac{dy}{dx}=-20x^{4}\)

\(g(x)=16x^{-2}\)

\(g'(x)=-32x^{-3}=-\frac{32}{x^{3}}\)

\(y=10(7-3)\)

\(y=40 \therefore y'=0\)

\(q(x)=x^{4}-6x^{2}-1\)

\(q'(x)=4x^{3}-12x\)

\(y=x^{2}+x+4\)

\(\frac{dy}{dx}=2x+1\)

\(f(x)=\frac{1}{3}x^{3} - x^{2} + \frac{2}{5}\)

\(f'(x)=x^{2}-2x\)

\(y=3x^{\frac{3}{2}} - 4x + 20\)

\(\frac{dy}{dx}=\frac{9}{2}x^{\frac{1}{2}}-4\)

\(g(x)=x(x+2)+5x\)

\(g(x)=x^{2}+7x \\ g'(x)=2x+7\)

\(p(x)=200[x^{3}-\frac{1}{2}x^{2}+\frac{1}{5}x-40]\)

\(p'(x)=200[3x^{2}-x+\frac{1}{5}] \\ p'(x)=600x^{2}-200x+40\)

\(y=14(x-1)\left[\frac{1}{2} + x^{2}\right]\)

\(y=14(x^{3} - x^{2} + \frac{1}{2}x - \frac{1}{2}) \\ \therefore \frac{dy}{dx} = 14\left(3x^{2} - 2x + \frac{1}{2}\right) \\ = 42x^{2}-28x+7\)

Find \({f}'\left(x\right)\) if \(f\left(x\right)=\frac{{x}^{2}-5x+6}{x-2}\).

\begin{align*} f(x) & = \frac{x^{2}-5x+6}{x-2}\\ & = \frac{(x-3)(x-2)}{x-2}\\ & = x-3\\ f'(x) & = \text{1} \end{align*}

Find \({f}'\left(y\right)\) if \(f\left(y\right)=\sqrt{y}\).

\begin{align*} f(y) & = \sqrt{y}\\ & = y^{\frac{1}{2}}\\ f'(y) & = \frac{1}{2}y^{-\frac{1}{2}}\\ & = \frac{1}{2\sqrt{y}} \end{align*}

Find \({f}'\left(z\right)\) if \(f\left(z\right)=\left(z-1\right)\left(z+1\right)\).

\begin{align*} f(z) & = (z-1)(z+1)\\ & = z^{2}-1\\ f'(z) & = 2z \end{align*}

Determine \(\frac{dy}{dx}\) if \(y=\frac{{x}^{3}+2\sqrt{x}-3}{x}\).

\begin{align*} y & = \frac{x^{3} + 2\sqrt{x}-3}{x}\\ y & = x^{2}+2x^{-\frac{1}{2}}-3x^{-1}\\ \frac{dy}{dx} & = 2x+2\left(-\frac{1}{2}\right)x^{-\frac{3}{2}}-3(-1)x^{-2}\\ & = 2x-\frac{1}{\sqrt{x^{3}}}+\frac{3}{x^{2}} \end{align*}

Determine the derivative of \(y=\sqrt{{x}^{3}}+\frac{1}{3{x}^{3}}\).

\begin{align*} y & = x^{\frac{3}{2}}+\frac{1}{3}x^{-3}\\ \frac{dy}{dx} & = \frac{3}{2}x^{\frac{1}{2}}-x^{-4}\\ & = \frac{3}{2}\sqrt{x} -\frac{1}{x^{4}} \end{align*}

Find \(D_{x}\left[x^{\frac{3}{2}} - \dfrac{3}{x^{\frac{1}{2}}}\right]^{2}\).

\begin{align*} D_{x}\left[x^{\frac{3}{2}} - \dfrac{3}{x^{\frac{1}{2}}}\right]^{2} & = D_{x}\left[x^{\frac{3}{2}} - \dfrac{3}{x^{\frac{1}{2}}}\right]\left[x^{\frac{3}{2}} - \dfrac{3}{x^{\frac{1}{2}}}\right]\\ & = D_{x}\left[x^{3}-2(3x)+\frac{9}{x}\right]\\ & = D_{x}\left[x^{3}-6x+9x^{-1}\right]\\ & = 3x^{2}-6-9x^{-2}\\ & = 3x^{2}-6-\frac{9}{x^{2}} \end{align*}

Find \(\frac{dy}{dx}\) if \(x=2y+3\).

Make \(y\) the subject of the formula in order to differentiate \(y\) with respect to \(x\).

\begin{align*} y & = \frac{1}{2}x - \frac{3}{2} \\ \therefore \frac{dy}{dx}& = \frac{1}{2} \end{align*}

Determine \(f'(\theta)\) if \(f(\theta) = 2(\theta^{\frac{3}{2}} - 3\theta^{-\frac{1}{2}})^{2}\).

\begin{align*} f(\theta) &= 2(\theta^{\frac{3}{2}} - 3\theta^{-\frac{1}{2}})^{2} \\ &= 2(\theta^{\frac{3}{2}} - 3\theta^{-\frac{1}{2}})(\theta^{\frac{3}{2}} - 3\theta^{-\frac{1}{2}}) \\ &= 2(\theta^{3} - 6\theta + 9\theta^{-1}) \\ \therefore f'(\theta) &= 2(3\theta^{2} - 6 - 9\theta^{-2}) \\ &= 2(3\theta^{2} - 6 - \frac{9}{\theta^{2}}) \\ &= 6\theta^{2} - 12 - \frac{18}{\theta^{2}} \end{align*}

Find \(\frac{dp}{dt}\) if \(p(t) = \frac{(t + 1)^{3}}{\sqrt{t}}\).

\begin{align*} p(t) &= \frac{(t + 1)^{3}}{\sqrt{t}} \\ &= \frac{t^{3} + 3t^{2} + 3t + 1}{t^{\frac{1}{2}}} \\ &= t^{-\frac{1}{2}}(t^{3} + 3t^{2} + 3t + 1) \\ &= t^{\frac{5}{2}} + 3t^{\frac{3}{2}} + 3t^{\frac{1}{2}} + t^{-\frac{1}{2}} \\ \therefore \frac{dp}{dt}&= \frac{5}{2}t^{\frac{3}{2}} + \frac{9}{2}t^{\frac{1}{2}} + \frac{3}{2}t^{-\frac{1}{2}} - \frac{1}{2}t^{-\frac{3}{2}} \\ &= \frac{5}{2}t^{\frac{3}{2}} + \frac{9}{2}t^{\frac{1}{2}} + \frac{3}{2t^{\frac{1}{2}}} - \frac{1}{2t^{\frac{3}{2}}} \end{align*}

6.2 Differentiation from first principles

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6.4 Equation of a tangent to a curve