Chain Rule | Product Rule | Quotient Rule | Sin Function | Cos Function | Tan Function | Exponential Function | Log Function |
\( y = f(x)^{n} \Rightarrow \frac{dy}{dx} = n \times f'(x) [f(x)]^{n} \)
\( y = uv \Rightarrow \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \)
\( y = \frac{u}{v} \Rightarrow \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^{2}} \)
\( y = \sin f(x) \Rightarrow \frac{dy}{dx} = f'(x) \cos f(x) \)
Step 1: Recognise this function is in the form \( \sin(f(x)) \) with \( f(x) = 3x^2 \).
Step 2: The derivative of \( f(x) = 3x^2 \) is \( f'(x) = 6x \).
Step 3: Multiply by \( \cos(f(x)) \): \( y' = \cos(3x^2) \times 6x = \ 6x \cos(3x^2) \)
\( y = \cos f(x) \Rightarrow \frac{dy}{dx} = - f'(x) \sin f(x) \)
Step 1: Recognise this function is in the form \( \cos(f(x)) \) with \( f(x) = 2x+1 \).
Step 2: The derivative of \( f(x) = 2x+1 \) is \( f'(x) = 2 \).
Step 3: Multiply by \( \cos(f(x)) \): \( y' = -\sin(2x+1) \times 2 = -2 \sin(2x+1) \)
\( y = \tan f(x) \Rightarrow \frac{dy}{dx} = f'(x) \sec^{2} f(x) \)
Step 1: Recognise this function is in the form \( \tan(f(x)) \) with \( f(x) = 5x \).
Step 2: The derivative of \( f(x) = 5x \) is \( f'(x) = 5 \).
Step 3: Multiply by \( \sec^{2}(f(x)) \): \( y' = \sec^{2}(5x) \times 5 = \ 5 \sec^{2}(5x) \)
\( y = e^{f(x)} \Rightarrow \frac{dy}{dx} = f'(x)e^{f(x)} \)
\( y = a^{f(x)} \Rightarrow \frac{dy}{dx} = ln(a) \times f'(x)a^{f(x)} \)
Step 1: Recognise this function is in the form \( e^{f(x)} \) with \( f(x) = x^{2}-3x-1 \).
Step 2: The derivative of \( f(x) = x^{2}-3x-1 \) is \( f'(x) = 2x - 3 \).
Step 3: Multiply by \( e^{f(x)} \): \( y' = e^{x^{2}-3x-1} \times (2x - 3) = \ (2x - 3) e^{x^{2}-3x-1} \)
\( y = \ln{f(x)} \Rightarrow \frac{dy}{dx} = \frac{f'(x)}{f(x)} \)
\( y = \log_{a}x \Rightarrow \frac{dy}{dx} = \frac{f'(x)}{( \ln a)f(x)} \)
Step 1: Recognise this function is in the form \( ln(f(x)) \) with \( f(x) = 4 - x^{3} \).
Step 2: The derivative of \( f(x) = 4 - x^{3} \) is \( f'(x) = -3x^{2}\).
Step 3: Multiply by \( \frac{1}{f(x)} \): \( y' = -3x^{2} \times \frac{1}{4 - x^{3}} = \ -\frac{3x^{2}}{4 - x^{3}} \)