What does it mean to differentiate a product of 2 functions? For example, for \[ u(x)= f(x).g(x) \ \] with \[ f(x) = x^4, \quad g(x) = x^2 \ \] A reasonable (but incorrect) first guess would be that the derivative of a product is the product of the derivatives.
You know from the power rule that \[ \begin{align} \frac{d}{dx} x^6 &= 6x^5 \end{align} \ \] so if you differentiate f and g and then multiply the results, you will end up short a power of x, regardless of the value of the coefficient. However, if you just differentiate f and leave g alone you will end up with the correct power of x. So \[ \begin{align} \frac{d}{dx} (x^4 \times x^2) &= 4x^3 \times x^2 + \text{something} \end{align} \ \] Now common sense suggests that f and g should play equal roles in the process, so the 'something' factor is f times 2x: \[ \begin{align} 6x^5 &= 4x^3 \times x^2 + x^4 \times 2x \\ \end{align} \ \] Observe that 2x is the derivative of g. So it looks as if \[ \begin{align} \frac{d}{dx} (f(x) \times g(x)) &= \frac{df}{dx} g(x) + f(x) \frac{dg}{dx} \quad \\ \end{align} \ \]
And you can plug these ideas into the limit definition of the derivative and eventually be able to prove that \[ \begin{align} \frac{d}{dx} f(x)g(x) &= f'(x)g(x) + f(x)g'(x) \end{align} \ \]
This is frequently called the product rule for differentiation.
Differentiate \( f(x) = 3 \times x^{4} \) using the product rule.
Differentiate \( f(x) = x^{1/2} \times x^{1/2} \) using the product rule.
Example Here is a slightly harder example \[ \begin{align} \frac{d}{dx} (3x^2\sqrt x) &= \frac{d}{dx}(3x^2)\times \sqrt x + 3x^2 \times \frac{d}{dx}(\sqrt x) \\ &= 6x\sqrt x + \frac{3x^2}{2\sqrt x} \\ &= \frac{6x2x + 3x^2}{2\sqrt x} \\ &= \frac{15x^2}{2\sqrt x} \end{align} \ \]
Guided Examples
O E Qs
When working exercises, it pays to be methodical. Here is an effective process, applied to \[ \begin{align} &\frac{d}{dx}(x^3+1)x^{1/2} \\ \end{align} \ \] First, plug the relevant parts of the expression into the formula: \[ \begin{align} &= \frac{d}{dx}(x^3+1) \times x^{1/2} + (x^3+1) \times \frac{d}{dx}(x^{1/2}) \quad \\ \end{align} \ \] Next, do the indicated differentiations: \[ \begin{align} &= 3x^2 \times x^{1/2} + (x^3+1) \times \frac{1}{2}(x^{-1/2}) \\ \end{align} \ \] Finally, simplify the result. \[ \begin{align} &= 3x^2x^{1/2} + \frac{x^3+1}{2x^{1/2}} \\ &= \frac{3x^22x + x^3 + 1}{2x^{1/2}} \\ &= \frac{7x^3 + 1}{2x^{1/2}} \end{align} \ \]
Differentiate \( f(x) = x \times x^{1/2} \) using the product rule.
In many cases you will need to use the chain rule to evaluate the derivative of one or both of the factors.
Example Extending the above example: \[ \begin{align} &\frac{d}{dx}(x^3+1)^3 x^{1/2} \\ \end{align} \ \] Plug in the expression \[ \begin{align} &= \frac{d}{dx}(x^3+1)^3 \times x^{1/2} + (x^3+1)^3 \times \frac{d}{dx}(x^{1/2}) \quad \\ \end{align} \ \] Do the differentiations. \[ \begin{align} &= 3(x^3+1)^2 \times 3x^2 \times x^{1/2} + (x^3+1)^3 \times \frac{1}{2}(x^{-1/2}) \\ \end{align} \ \] Simplify \[ \begin{align} &= (x^3+1)^2 9x^2 x^{1/2} + (x^3 + 1)\frac{1}{2x^{1/2}} \\ &= (x^3 + 1)[(x^3+1)9x^2 x^{1/2} + \frac{1}{2x^{1/2}}] \\ &= (x^3 + 1)[\frac{(x^3+1)9x^2 2x + 1}{2x^{1/2}}] \\ &= (x^3 + 1)[\frac{18x^6 + 18x^3 + 1}{2x^{1/2}}] \\ \end{align} \ \]
Differentiate \( f(x) = \sqrt{x^2+1} \times x^3 \) using the product rule.