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Derivatives: Trigonometric functions |
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In this section, we will cover the aspects of taking derivatives of trigonometric functions
Derivatives of Trigonometric Functions
=cos(x) "\frac{d}{dx} sin(x)=cos(x)")
=-sin(x) "\frac{d}{dx} cos(x)=-sin(x)")
=sec^2(x) "\frac{d}{dx} tan(x)=sec^2(x)")
=-csc(x)cot(x) "\frac{d}{dx} csc(x)=-csc(x)cot(x)")
=sec(x)tan(x) "\frac{d}{dx} sec(x)=sec(x)tan(x)")
=-csc^2(x) "\frac{d}{dx} cot(x)=-csc^2(x)")
Examples
Find the derivatives of the following functions
=sin(x)+cos(x) "1. \,\,f(x)=sin(x)+cos(x)")
=cos(x)-sin(x) "f\prime(x)=cos(x)-sin(x)")
This completes example 1
=sin^2(x) = (sin(x))^2 "2. \,\, f(x)=sin^2(x) = (sin(x))^2")
The derivative of this function may be found in two ways, by using the product rule or chain rule. We will use both ways:
Using the product rule:
=sin(x)sin(x) "f(x)=sin(x)sin(x)")
=sin(x)cos(x)+cos(x)sin(x)= 2sin(x)cos(x) "f\prime(x)=sin(x)cos(x)+cos(x)sin(x)= 2sin(x)cos(x)")
And using the chain rule
=(sin(x))^2 "f(x)=(sin(x))^2")
=2(sin(x))cos(x) "f\prime(x)=2(sin(x))cos(x)")
We can see that either way produces the same result. This completes example 2
=\frac{1}{csc(x)} "3. \,\, f(x)=\frac{1}{csc(x)}")
This is just the same as
=(csc(x))^{-1} "f(x)=(csc(x))^{-1}")
Using the chain rule,
=-1(csc(x))^{-2} = \frac{-1}{csc^2(x)} "f\prime(x)=-1(csc(x))^{-2} = \frac{-1}{csc^2(x)}")
This completes example 3
=sin(2x)cos(5x) "4. \,\, f(x)=sin(2x)cos(5x)")
In this example, we will employ both the chain rule and product rules
=(2cos(2x))(cos(5x))+(sin(2x))(-5sin(5x)) = 2cos(2x)cos(5x)-5sin(2x)cos(5x) "f\prime(x)=(2cos(2x))(cos(5x))+(sin(2x))(-5sin(5x)) = 2cos(2x)cos(5x)-5sin(2x)cos(5x)")
This completes example 4
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