Rules For Differentiation
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.
Optional Investigation
Rules for differentiation
Differentiate the following from first principles:
\(f(x)=x\)
\(f(x)=-4x\)
\(f(x)={x}^{2}\)
\(f(x)=3{x}^{2}\)
\(f(x)=-{x}^{3}\)
\(f(x)=2{x}^{3}\)
\(f(x)=\cfrac{1}{x}\)
\(f(x)=-\cfrac{2}{x}\)
Complete the table:
\(f(x)\) \({f}'(x)\) \(x\) \(-4x\) \(x^{2}\) \(3x^{2}\) \(-x^{3}\) \(2x^{3}\) \(\cfrac{1}{x}\) \(-\cfrac{2}{x}\) - Can you identify a pattern for determining the derivative?
Rules for Differentiation
General rule for differentiation:
\[\cfrac{d}{dx}[{x}^{n}]=n{x}^{n-1}, \text{ where } n \in \mathbb{R} \text{ and } n \ne 0.\]
The derivative of a constant is equal to zero.
\[\cfrac{d}{dx}[k]= 0\]
The derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function.
\[\cfrac{d}{dx}[k \cdot f(x) ]=k \cfrac{d}{dx}[ f(x) ]\]
The derivative of a sum is equal to the sum of the derivatives.
\[\cfrac{d}{dx}[f(x)+g(x)]=\cfrac{d}{dx}[f(x) ] + \cfrac{d}{dx}[g(x)]\]
The derivative of a difference is equal to the difference of the derivatives.
\[\cfrac{d}{dx}[f(x) – g(x)]=\cfrac{d}{dx}[f(x) ] – \cfrac{d}{dx}[g(x)]\]
Example
Question
Use the rules of differentiation to find the derivative of each of the following:
- \(y = 3x^{5}\)
- \(p = \cfrac{1}{4}q^{2}\)
- \(f(x) = 60\)
- \(y = 12x^{3} + 7x\)
- \(m = \cfrac{3}{2}n^{4} – 1\)
Apply the appropriate rules to determine the derivative
- \(\cfrac{dy}{dx} = 3 (5x^{4}) = 15x^{4}\)
- \(\cfrac{dp}{dq} = \cfrac{1}{4}(2q) = \cfrac{1}{2}q\)
- \(f'(x) = 0\)
- \(\cfrac{dy}{dx} = 12(3x^{2}) + 7 = 36x^{2} + 7\)
- \(\cfrac{dm}{dn} = \cfrac{3}{2}(4n^{3}) – 0 = 6n^{3}\)
Example
Question
Differentiate the following with respect to \(t\):
- \(g(t) = 4( t + 1 )^{2} ( t -3 )\)
- \(k(t) = \cfrac{(t + 2)^{3}}{\sqrt{t}}\)
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( t + 1 )^{2} ( t – 3 ) \\ &= 4( t^{2} + 2t + 1 ) ( t – 3 ) \\ &= 4( t^{3} + 2t^{2} + t – 3t^{2} – 6t – 3 ) \\ &= 4( t^{3} – t^{2} – 5t – 3 ) \\ &= 4t^{3} – 4t^{2} – 20t – 12 \\ & \\ \therefore g'(t) &= 4 ( 3t^{2} ) – 4( 2t ) – 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) &= \cfrac{(t + 2)^{3}}{\sqrt{t}} \\ &= \cfrac{(t + 2)(t^{2} + 4t + 4)}{\sqrt{t}} \\ &= \cfrac{ t^{3} + 6t^{2} + 12t + 8}{t^{\frac{1}{2}}} \\ &= t^{-\cfrac{1}{2}} ( t^{3} + 6t^{2} + 12t + 8 ) \\ &= t^{\frac{5}{2}} + 6t^{\frac{3}{2}} + 12t^{\frac{1}{2}} + 8t^{-\cfrac{1}{2}} \\ & \\ \therefore g'(t) &= \cfrac{5}{2}t^{\frac{3}{2}} + 6 ( \cfrac{3}{2}t^{\frac{1}{2}} ) + 12 ( \cfrac{1}{2} t^{-\cfrac{1}{2}} ) + 8 ( – \cfrac{1}{2} t^{-\cfrac{3}{2}} ) \\ &= \cfrac{5}{2}t^{\frac{3}{2}} + 9t^{\frac{1}{2}} + 6t^{-\cfrac{1}{2}} – 4t^{-\cfrac{3}{2}} \end{align*}
Important: always write the final answer with positive exponents.
\begin{align*} g'(t) &= \cfrac{5}{2}t^{\frac{3}{2}} + 9t^{\frac{1}{2}} + \cfrac{6}{t^{\frac{1}{2}}} – \cfrac{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.