Trigonometric Derivatives, as with all derivatives, are used to find the rate of change of trigonometric functions. They are used in various real-world applications such as programming and electronics.
Below is a table outlining the derivatives of each trig function:
Function |
Original |
Derivative |
Sine |
\(f(x) = \sin(x)\) |
\(f'(x) = \cos(x)\) |
Cosine |
\(f(x) = \cos(x)\) |
\(f'(x) = -\sin(x)\) |
Tangent |
\(f(x) = \tan(x)\) |
\(f'(x) = \cfrac{1}{\text{cos}^2x} = \sec^2(x)\) |
Cotangent |
\(f(x) = \cot(x)\) |
\(f'(x) = -\cfrac{1}{\text{sin}^2x} = -\csc^2(x)\) |
Secant |
\(f(x) = \sec(x)\) |
\(f(x)' = \tan(x)\sec(x)\) |
Cosecant |
\(f'(x) = \csc(x)\) |
\(f'(x) = -\cot(x)\csc(x)\) |
Example
Differentiate the following equations.
- \(y = \cos(3x)\)
- \(y = \tan(x)\sin(2x)\)
i. We can use Chain Rule to differentiate this equation:
\(y' = \textcolor{red}{f'(g(x))}\textcolor{blue}{g'(x)}\)
\(y' = \textcolor{red}{-\sin(3x)}\cdot \textcolor{blue}{3}\)
\(y' = -3\sin((3x))\)
Therefore, we can determine that \(y' = -3\sin(3x)\).
ii. We can use Product Rule to find the derivative of this expression:
First, we can identify \(f(x)\) and \(g(x)\):
\(\textcolor{red}{f(x) = \tan(x)}\)
\(\textcolor{blue}{g(x) = \sin(2x)}\)
We can determine \(f'(x)\) simply by differentiating \(f(x)\):
\(\textcolor{red}{f'(x) = \sec^2(x)}\)
We can determine \(g'(x)\) by using Chain Rule:
\(g'(x) = \cos(2x)\cdot 2\)
\(\textcolor{blue}{g'(x) = 2\cos(2x)}\)
Next, we can use Product Rule to differentiate the function:
\(y' = \textcolor{red}{f'(x)}\textcolor{blue}{g(x)} + \textcolor{blue}{g'(x)}\textcolor{red}{f(x)}\)
\(y' = \textcolor{red}{\sec^2(x)}\textcolor{blue}{\sin(2x)} + \textcolor{blue}{2\cos(2x)}\textcolor{red}{\tan(x)}\)
Therefore, we can determine that \(y' = \sec^2(x)\sin(2x) + 2\cos(2x)\tan(x)\).
Differentiate the following equations.
\(y = \cos^3(x^2+\pi x)\)
Show Answer
First, we can determine \(f'(g(x))\). This involves using Chain Rule twice. We can use it to multiply the original function by its exponent (\(3\)):
\(= 3\cos^2(x^2+\pi x)\)
We can then use it multiply the function by its derivative trigonometric function \(-\sin\):
\(= -\sin(x^2+\pi x)\)
When we put these 2 portions together, we get \(f'(g(x))\):
\(\textcolor{red}{f'(x) = 3\cos^2(x^2+\pi x)[-\sin(x^2+\pi x)]}\)
Next, we can determine \(g'(x)\):
\(\textcolor{blue}{g'(x) = 2x + \pi}\)
Then, we can implement Chain Rule to determine the derivative:
\(y' = \textcolor{red}{f'(g(x))}\textcolor{blue}{g'(x)}\)
\(y' = \textcolor{red}{3\cos (x)^2((x)^2+\pi x)[-\sin(x^2+\pi (x)]}\textcolor{blue}{[2(x)+\pi]}\)
We cannot simplify this function any further.
Therefore, we can determine that \(y' = 3\cos (x)^2(x^2+\pi (x))[-\sin((x)^2+\pi (x)][2(x)+\pi]\).
\(y = 2\sin^3(x)-4\cos^2(x)\)
Show Answer
We can use the Difference Rule to differentiate the expression.
First, we can determine \(f'(x)\):
\(f'(x) = 3(2\sin^2(x))\cdot \cos(x)\)
\(\textcolor{red}{f'(x) = 6\sin^2(x)\cos(x)}\)
Next, we can determine \(h'(x)\):
\(g'(x) = -2(4\cos(x)) \cdot (-\sin(x))\)
\(\textcolor{blue}{g'(x) = 8\cos(x)\sin(x)}\)
Then, we can implement the Sum/Difference Rule to find the derivative of this expression:
\(y' = \textcolor{red}{f'(x)} + \textcolor{blue}{g'(x)}\)
\(y' = \textcolor{red}{6\sin^2(x)\cos(x)} - \textcolor{blue}{8\cos(x)\sin(x)}\)
Finally, we can simplify the expression further by taking a common factor out of both terms:
\(y' = [\cos (x)\sin (x)][6\sin (x) + 8]\)
Therefore, we can determine that \(y' = [\cos (x)\sin (x)][8\sin (x) + 8]\).