Apply Differentiation Rules (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example 1: Product rule with chain rule

Question: Calculate the derivative of $y = (5x^3 - 8)^2\left(\frac{x}{2} - 2\right)$.

Strategy: This function is a product of two expressions, so the product rule is the primary rule. The first term, $(5x^3 - 8)^2$, will need the chain rule for differentiation.

Solution:

Let $u = (5x^3 - 8)^2$ and $v = \frac{x}{2} - 2$.

For $\frac{du}{dx}$:

$u = (5x^3 - 8)^2$

$\frac{du}{dx} = 2(5x^3 - 8) \times \frac{d}{dx}(5x^3 - 8)$

$= 2(5x^3 - 8) \times (15x^2)$

$= 30x^2(5x^3 - 8)$

For $\frac{dv}{dx}$:

$v = \frac{x}{2} - 2$

$\frac{dv}{dx} = \frac{1}{2}$

Applying the product rule:

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

$= (5x^3 - 8)^2\left(\frac{1}{2}\right) + \left(\frac{x}{2} - 2\right)(30x^2)(5x^3 - 8)$

$= (5x^3 - 8)\left[(5x^3 - 8)\left(\frac{1}{2}\right) + \left(\frac{x}{2} - 2\right)(30x^2)\right]$

$= (5x^3 - 8)\left[\frac{5}{2}x^3 - 4 + 15x^3 - 60x^2\right]$

$= (5x^3 - 8)\left(\frac{35}{2}x^3 - 60x^2 - 4\right)$

Worked Example 2: Quotient rule with power rule

Question: Calculate the gradient of the graph of $y = \sqrt{\frac{x^2 + 9}{2x^3 + 1}}$ at the point where $x = 0$.

Strategy: Use the power rule formula $y = (f(x))^n$ where $y' = n(f(x))^{n-1}f'(x)$, then evaluate the gradient at $x = 0$.

Solution:

Let $y = f(x) = \frac{x^2 + 9}{2x^3 + 1}$.

For $f'(x)$:

$f'(x) = \frac{(2x^3 + 1)(2x) - (x^2 + 9)(6x^2)}{(2x^3 + 1)^2}$

$= \frac{(4x^4 + 2x) - (6x^4 + 54x^2)}{(2x^3 + 1)^2}$

$= \frac{-2x^4 - 54x^2 + 2x}{(2x^3 + 1)^2}$

For $y'$:

$y = \sqrt{\frac{x^2 + 9}{2x^3 + 1}}$

$= \left(\frac{x^2 + 9}{2x^3 + 1}\right)^{\frac{1}{2}}$

$y' = \frac{1}{2}\left(\frac{x^2 + 9}{2x^3 + 1}\right)^{-\frac{1}{2}} \times \frac{-2x^4 - 54x^2 + 2x}{(2x^3 + 1)^2}$

$= \frac{1}{2\sqrt{\frac{x^2 + 9}{2x^3 + 1}}} \times \frac{-2x^4 - 54x^2 + 2x}{(2x^3 + 1)^2}$

Calculate the gradient at $x = 0$:

$y' = \frac{1}{2\sqrt{\frac{0^2 + 9}{2(0)^3 + 1}}} \times \frac{0 - 0 + 0}{(0 + 1)^2}$

$= \frac{1}{2\sqrt{\frac{9}{1}}} \times \frac{0}{1}$

$= \frac{1}{2\sqrt{9}} \times \frac{0}{1}$

$= \frac{1}{6} \times 0$

$= 0$

The gradient at $x = 0$ is $0$.

Worked Example 3: Product rule for tangent line

Question: Determine the equation of the tangent line to the graph of $f(x) = 2\sqrt{x}(x + 8)(4 - x^2)$ at the point where $x = 1$.

Strategy: The function is primarily a product where one factor is itself a product. Apply the product rule to find the derivative, substitute $x = 1$ to find the gradient, then use the point-gradient formula to find the tangent equation.

Solution:

Let $u = 2(\sqrt{x})(x + 8)$.

$u = 2\sqrt{x}(x + 8)$

$= 2x^{\frac{1}{2}}(x + 8)$

$= 2x^{\frac{3}{2}} + 16x^{\frac{1}{2}}$

So $u = 2x^{\frac{3}{2}} + 16x^{\frac{1}{2}}$ and $v = 4 - x^2$.

For $u'$:

$u = 2x^{\frac{3}{2}} + 16x^{\frac{1}{2}}$

$u' = 3x^{\frac{1}{2}} + 8x^{-\frac{1}{2}}$

For $v'$:

$v = 4 - x^2$

$v' = -2x$

Applying the product rule $f'(x) = uv' + vu'$:

$f'(x) = uv' + vu'$

$= \left(2x^{\frac{3}{2}} + 16x^{\frac{1}{2}}\right)(-2x) + (4 - x^2)\left(3x^{\frac{1}{2}} + 8x^{-\frac{1}{2}}\right)$

Calculate the gradient $m = f'(x)$ by substituting $x = 1$:

$m = f'(1)$

$= \left(2(1)^{\frac{3}{2}} + 16(1)^{\frac{1}{2}}\right)(-2(1)) + (4 - 1^2)\left(3(1)^{\frac{1}{2}} + 8(1)^{-\frac{1}{2}}\right)$

$= \left(2(1) + 16(1)\right)(-2(1)) + (4 - 1)\left(3(1) + 8(1)\right)$

$= (2 + 16)(-2) + (3)(3 + 8)$

$= (18) \times (-2) + (3) \times (11)$

$= -36 + 33$

$= -3$

Now substitute $x = 1$ into the function to calculate the $y$-coordinate:

$f(x) = 2\sqrt{x}(x + 8)(4 - x^2)$

$f(1) = 2\sqrt{1}(1 + 8)(4 - 1^2)$

$= 2(1)(9)(3)$

$= 54$

The point of tangency is at $(1, 54)$.

Using the point-gradient formula with gradient $m = -3$ and point $(1, 54)$:

$y - y_1 = m(x - x_1)$

$y - 54 = -3(x - 1)$

$y - 54 = -3x + 3$

$y = -3x + 57$

The equation of the tangent line is $y = -3x + 57$.