Trigonometric derivatives are used to find the rate of change of functions involving sine, cosine, and other trig functions. This page focuses on basic sine and cosine derivative rules, along with product rule, quotient rule, and chain rule variations. These examples help build fluency with common Calculus 1 trig derivative problems.
Notes

Problems
Find the derivative of the function.
\(\textbf{1)}\) \(f(x)=-3 \cos x\)
The derivative is \(f'(x)=3 \sin x\)
\(\,\,\,\,\,f(x)=-3\cos x\)
\(\,\,\,\,\,\frac{d}{dx}\left(\cos x\right)=-\sin x\)
\(\,\,\,\,\,f'(x)=-3(-\sin x)\)
\(\,\,\,\,\,f'(x)=3\sin x\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=3 \sin x\)
\(\textbf{2)}\) \(f(x)=\pi-\cos x\)
The derivative is \(f'(x)=\sin x\)
\(\,\,\,\,\,f(x)=\pi-\cos x\)
\(\,\,\,\,\,\frac{d}{dx}\left(\pi\right)=0\)
\(\,\,\,\,\,\frac{d}{dx}\left(-\cos x\right)=\sin x\)
\(\,\,\,\,\,f'(x)=\sin x\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\sin x\)
\(\textbf{3)}\) \(f(x) = \pi \sin x + 2\pi x\)
The derivative is \(f'(x) = \pi \cos x + 2\pi\)
\(\,\,\,\,\,f(x) = \pi \sin x + 2\pi x\)
\(\,\,\,\,\,\frac{d}{dx}\left(\sin x\right)=\cos x\)
\(\,\,\,\,\,\frac{d}{dx}\left(x\right)=1\)
\(\,\,\,\,\,f'(x) = \pi \cdot \cos x + 2\pi(1)\)
\(\,\,\,\,\,f'(x) = \pi \cos x + 2\pi\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = \pi \cos x + 2\pi\)
\(\textbf{4)}\) \(f(x)=-4 \sin x+4 \cos x\)
The derivative is \(f'(x)=-4 \cos x-4 \sin x\)
\(\,\,\,\,\,f(x) = -4 \sin x + 4 \cos x\)
\(\,\,\,\,\,\frac{d}{dx}\left(\sin x\right)=\cos x\)
\(\,\,\,\,\,\frac{d}{dx}\left(\cos x\right)=-\sin x\)
\(\,\,\,\,\,f'(x) = -4 \cdot \cos x + 4 \cdot (-\sin x)\)
\(\,\,\,\,\,f'(x) = -4 \cos x – 4 \sin x\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=-4 \cos x-4 \sin x\)
\(\textbf{5)}\) \(f(x)=5x^{3} \cos x\)
The derivative is \(f'(x)=-5x^{3} \sin x+15x^{2} \cos x\)
\(\,\,\,\,\,\text{This will use product rule}\)
\(\,\,\,\,\,f(x) = 5x^{3} \cos x\)
\(\,\,\,\,\,f'(x) = (5x^{3}) \cdot (\cos x)’ + (5x^{3})’ \cdot (\cos x)\)
\(\,\,\,\,\,f'(x) = 5x^{3} \cdot (-\sin x) + 15x^{2} \cdot \cos x\)
\(\,\,\,\,\,f'(x) = -5x^{3} \sin x + 15x^{2} \cos x\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=-5x^{3} \sin x+15x^{2} \cos x\)
\(\textbf{6)}\) \(f(x) = 4 x^2 \sin (x) \)
The derivative is \(f'(x) = 4x^2 \cos x + 8x \sin x\)
\(\,\,\,\,\,\text{This will use product rule}\)
\(\,\,\,\,\,f(x) = 4x^2 \sin x\)
\(\,\,\,\,\,f'(x) = 4x^2(\sin x)’ + (4x^2)’\sin x\)
\(\,\,\,\,\,f'(x) = 4x^2 \cos x + 8x \sin x\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = 4x^2 \cos x + 8x \sin x\)
\(\textbf{7)}\) \(f(x) = \cos (x) \sin (x)\)
The derivative is \(f'(x) = \cos^2 x – \sin^2 x\)
\(\,\,\,\,\,\text{This will use product rule}\)
\(\,\,\,\,\,f(x) = \cos x \cdot \sin x\)
\(\,\,\,\,\,f'(x) = (\cos x) \cdot (\sin x)’ + (\sin x) \cdot (\cos x)’\)
\(\,\,\,\,\,f'(x) = \cos x \cos x – \sin x \sin x\)
\(\,\,\,\,\,f'(x) = \cos^2 x – \sin^2 x\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = \cos^2 x – \sin^2 x\)
\(\textbf{8)}\) \(f(x) = \sin(3x^2)\)
The derivative is \(f'(x) = 6x \cos(3x^2)\)
\(\,\,\,\,\,\text{This will use chain rule}\)
\(\,\,\,\,\,f(x) = \sin(3x^2)\)
\(\,\,\,\,\,f'(x) = \cos(3x^2) \cdot (3x^2)’\)
\(\,\,\,\,\,f'(x) = \cos(3x^2) \cdot 6x\)
\(\,\,\,\,\,f'(x) = 6x \cos(3x^2)\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = 6x \cos(3x^2)\)
\(\textbf{9)}\) \(f(x) = \displaystyle \frac{\sin x}{x^2}\)
The derivative is \(f'(x) = \frac{x \cos x – 2 \sin x}{x^3}\)
\(\,\,\,\,\,\text{This will use quotient rule}\)
\(\,\,\,\,\,f(x) = \frac{\sin x}{x^2}\)
\(\,\,\,\,\,f'(x) = \frac{(\sin x)’ \cdot x^2 – \sin x \cdot (x^2)’}{(x^2)^2}\)
\(\,\,\,\,\,f'(x) = \frac{x^2 \cos x – 2 x \sin x}{x^4}\)
\(\,\,\,\,\,f'(x) = \frac{x\left(x \cos x – 2 \sin x\right)}{x^4}\)
\(\,\,\,\,\,f'(x) = \frac{x \cos x – 2 \sin x}{x^3}\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = \frac{x \cos x – 2 \sin x}{x^3}\)
\(\textbf{10)}\) \(f(x) = x^3 \sin(x^2)\)
The derivative is \(f'(x) = 3x^2 \sin(x^2) + 2x^4 \cos(x^2)\)
\(\,\,\,\,\,f(x) = x^3 \sin(x^2)\)
\(\,\,\,\,\,\text{This will use product rule and chain rule}\)
\(\,\,\,\,\,f'(x) = (x^3)’ \cdot \sin(x^2) + x^3 \cdot (\sin(x^2))’\)
\(\,\,\,\,\,f'(x) = 3x^2 \sin(x^2) + x^3 \cos(x^2) \cdot 2x\)
\(\,\,\,\,\,f'(x) = 3x^2 \sin(x^2) + 2x^4 \cos(x^2)\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = 3x^2 \sin(x^2) + 2x^4 \cos(x^2)\)
\(\textbf{11)}\) \(f(x) = \sin^2 x\)
The derivative is \(f'(x) = 2 \sin x \cos x\)
\(\,\,\,\,\,\text{This will use chain rule}\)
\(\,\,\,\,\,f(x) = \sin^2 x\)
\(\,\,\,\,\,f(x) = \left( \sin x \right)^2\)
\(\,\,\,\,\,f'(x) = 2 \sin x \cdot \left(\sin x \right)’\)
\(\,\,\,\,\,f'(x) = 2 \sin x \cdot \cos x\)
\(\,\,\,\,\,f'(x) = 2 \sin x \cos x\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = 2 \sin x \cos x\)
\(\textbf{12)}\) \(f(x) = x^2 \cos(2x)\)
The derivative is \(f'(x) = 2x \cos(2x) – 2x^2 \sin(2x)\)
\(\,\,\,\,\,\text{This will use product rule and chain rule}\)
\(\,\,\,\,\,f(x) = x^2 \cos(2x)\)
\(\,\,\,\,\,f'(x) = (x^2)’ \cdot \cos(2x) + x^2 \cdot (\cos(2x))’\)
\(\,\,\,\,\,f'(x) = 2x \cos(2x) + x^2 \cdot \left(-\sin(2x)\right) \cdot 2\)
\(\,\,\,\,\,f'(x) = 2x \cos(2x) – 2x^2 \sin(2x)\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = 2x \cos(2x) – 2x^2 \sin(2x)\)
\(\textbf{13)}\) \(f(x) = \displaystyle \frac{\cos x}{x}\)
The derivative is \(f'(x) = \frac{-x \sin x – \cos x}{x^2}\)
\(\,\,\,\,\,\text{This will use quotient rule}\)
\(\,\,\,\,\,f(x) = \frac{\cos x}{x}\)
\(\,\,\,\,\,f'(x) = \frac{(\cos x)’ \cdot x – \cos x \cdot (x)’}{x^2}\)
\(\,\,\,\,\,f'(x) = \frac{ \left(-\sin x \right) \cdot x – \cos x \cdot (1)}{(x)^2}\)
\(\,\,\,\,\,f'(x) = \frac{-x \sin x – \cos x}{x^2}\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = \frac{-x \sin x – \cos x}{x^2}\)
\(\textbf{14)}\) \(f(x) = x \sin(5x)\)
The derivative is \(f'(x) = \sin(5x) + 5x \cos(5x)\)
\(\,\,\,\,\,\text{This will use product rule}\)
\(\,\,\,\,\,f(x) = x \sin(5x)\)
\(\,\,\,\,\,f'(x) = (x)’ \cdot \sin(5x) + x \cdot (\sin(5x))’\)
\(\,\,\,\,\,f'(x) = (1) \cdot \sin(5x) + x \cdot (\cos(5x)) \cdot 5\)
\(\,\,\,\,\,f'(x) = \sin(5x) + 5x \cos(5x)\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = \sin(5x) + 5x \cos(5x)\)
\(\textbf{15)}\) \(f(x) = \displaystyle \frac{x^3}{\sin x}\)
The derivative is \(f'(x) = \frac{3x^2 \sin x – x^3 \cos x}{\sin^2 x}\)
\(\,\,\,\,\,\text{This will use quotient rule}\)
\(\,\,\,\,\,f(x) = \frac{x^3}{\sin x}\)
\(\,\,\,\,\,f'(x) = \frac{(x^3)’ \cdot \sin x – x^3 \cdot (\sin x)’}{(\sin x)^2}\)
\(\,\,\,\,\,f'(x) = \frac{3x^2 \sin x – x^3 \cos x}{\sin^2 x}\)
\(\,\,\,\,\,\)The derivative is \(f'(x) = \frac{3x^2 \sin x – x^3 \cos x}{\sin^2 x}\)
\(\textbf{16)}\) \(f(x)=7\sin x-9\cos x\)
The derivative is \(f'(x)=7\cos x+9\sin x\)
\(\,\,\,\,\,f(x)=7\sin x-9\cos x\)
\(\,\,\,\,\,\frac{d}{dx}\left(\sin x\right)=\cos x\)
\(\,\,\,\,\,\frac{d}{dx}\left(\cos x\right)=-\sin x\)
\(\,\,\,\,\,f'(x)=7\cos x-9(-\sin x)\)
\(\,\,\,\,\,f'(x)=7\cos x+9\sin x\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=7\cos x+9\sin x\)
\(\textbf{17)}\) \(f(x)=\cos(4x^3)\)
The derivative is \(f'(x)=-12x^2\sin(4x^3)\)
\(\,\,\,\,\,f(x)=\cos(4x^3)\)
\(\,\,\,\,\,\text{This will use chain rule}\)
\(\,\,\,\,\,u=4x^3\)
\(\,\,\,\,\,u’=12x^2\)
\(\,\,\,\,\,f'(x)=-\sin(4x^3)\cdot 12x^2\)
\(\,\,\,\,\,f'(x)=-12x^2\sin(4x^3)\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=-12x^2\sin(4x^3)\)
\(\textbf{18)}\) \(f(x)=\left(x^2+1\right)\cos x\)
The derivative is \(f'(x)=2x\cos x-\left(x^2+1\right)\sin x\)
\(\,\,\,\,\,f(x)=\left(x^2+1\right)\cos x\)
\(\,\,\,\,\,\text{This will use product rule}\)
\(\,\,\,\,\,f'(x)=\left(x^2+1\right)(\cos x)’+\left(x^2+1\right)’\cos x\)
\(\,\,\,\,\,f'(x)=\left(x^2+1\right)(-\sin x)+2x\cos x\)
\(\,\,\,\,\,f'(x)=2x\cos x-\left(x^2+1\right)\sin x\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=2x\cos x-\left(x^2+1\right)\sin x\)
\(\textbf{19)}\) \(f(x)=\displaystyle\frac{x^2}{\cos x}\)
The derivative is \(f'(x)=\displaystyle\frac{2x\cos x+x^2\sin x}{\cos^2 x}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{x^2}{\cos x}\)
\(\,\,\,\,\,\text{This will use quotient rule}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{(x^2)’\cos x-x^2(\cos x)’}{\left(\cos x\right)^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{2x\cos x-x^2(-\sin x)}{\cos^2 x}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{2x\cos x+x^2\sin x}{\cos^2 x}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{2x\cos x+x^2\sin x}{\cos^2 x}\)
\(\textbf{20)}\) \(f(x)=\sin x \cos(2x)\)
The derivative is \(f'(x)=\cos x\cos(2x)-2\sin x\sin(2x)\)
\(\,\,\,\,\,f(x)=\sin x \cos(2x)\)
\(\,\,\,\,\,\text{This will use product rule and chain rule}\)
\(\,\,\,\,\,f'(x)=\sin x\left(\cos(2x)\right)’+(\sin x)’\cos(2x)\)
\(\,\,\,\,\,f'(x)=\sin x\left(-2\sin(2x)\right)+\cos x\cos(2x)\)
\(\,\,\,\,\,f'(x)=\cos x\cos(2x)-2\sin x\sin(2x)\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\cos x\cos(2x)-2\sin x\sin(2x)\)
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