Derivative of Trigonometric Functions
Derivative of Sine and Cosine Function
There are two very important limits we must know before we discuss the derivative of trig functions. See below:
Important:
[latex]\Large{\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1}[/latex]
[latex]\Large{\lim_{\theta \to 0}\frac{\cos\theta -1}{\theta} = 0}[/latex]
It is important to remember that for the above limits to be true, [latex]\theta[/latex] must be measured in radians and not degrees. So, when dealing with trigonometric functions in calculus, ensure to keep the measure of the angles in radians.
Now, what is the derivative of sin(x)? Let us use the definition of the derivative to find this, as shown below:
[latex]\large{\frac{d}{dx}\sin(x) = \lim_{h \rightarrow 0}\frac{\sin(x+h) – \sin(x)}{h}}[/latex]
As we know, we cannot just substitute h = 0 to solve the limit, so we must use the following trig identity to first rewrite the numerator. We know that:
[latex]\large{\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)}[/latex]
Hence, we have:
[latex]\large{\frac{d}{dx}\sin(x) = \lim_{h \rightarrow 0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}}[/latex]
[latex]\large{= \lim_{h \rightarrow 0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}}[/latex]
[latex]\large{= \lim_{h \rightarrow 0}sin(x)\frac{\cos(h)-1}{h}+\lim_{h \rightarrow 0}\cos(x)\frac{\sin(h)}{h}}[/latex]
[latex]\large{= \sin(x)\lim_{h \rightarrow 0}\frac{\cos(h)-1}{h}+\cos(x)\lim_{h \rightarrow 0}\frac{\sin(h)}{h}}[/latex]
Now, we have already determined the limits of [latex]\frac{\cos(h) – 1}{h}[/latex] and [latex]\frac{\sin(h)}{h}[/latex] as h approaches zero. Using these definitions, we now have:
[latex]\large{\frac{d}{dx}\sin(x) = \sin(x)(0) + \cos(x)(1) = \cos(x)}[/latex]
This shows us that the derivative of sin(x) is cos(x). In a similar manner, we can show that the derivative of cos(x) is -sin(x). Try it out yourself!
It is important to remember that for the above limits to be true, [latex]\theta[/latex] must be measured in radians and not degrees. So, when dealing with trigonometric functions in calculus, ensure to keep the measure of the angles in radians.
Now, what is the derivative of sin(x)? Let us use the definition of the derivative to find this, as shown below:
[latex]\frac{d}{dx}\sin(x) = \lim_{h \rightarrow 0}\frac{\sin(x+h) – \sin(x)}{h}[/latex]
As we know, we cannot just substitute h = 0 to solve the limit, so we must use the following trig identity to first rewrite the numerator. We know that:
[latex]\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)[/latex]
Hence, we have:
[latex]\frac{d}{dx}\sin(x) = \lim_{h \rightarrow 0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}[/latex]
[latex]= \lim_{h \rightarrow 0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}[/latex]
[latex]= \small{\lim_{h \rightarrow 0}\sin(x)\frac{\cos(h)-1}{h}+\lim_{h \rightarrow 0}\cos(x)\frac{\sin(h)}{h}}[/latex]
[latex]= \small{\sin(x)\lim_{h \rightarrow 0}\frac{\cos(h)-1}{h}+\cos(x)\lim_{h \rightarrow 0}\frac{\sin(h)}{h}}[/latex]
Now, we have already determined the limits of [latex]\frac{\cos(h) – 1}{h}[/latex] and [latex]\frac{\sin(h)}{h}[/latex] as h approaches zero. Using these definitions, we now have:
[latex]\frac{d}{dx}\sin(x) = \sin(x)(0) + \cos(x)(1) = \cos(x)[/latex]
This shows us that the derivative of sin(x) is cos(x). In a similar manner, we can show that the derivative of cos(x) is -sin(x). Try it out yourself!
Example
Using the definition of the derivative, show that:
[latex]\large{\frac{d}{dx}cos(x) = -sin(x)}[/latex]
Step 1:
Substituting into the definition of the derivative, we have:
[latex]\large{\frac{d}{dx}(cos(x)) = \lim_{h\to 0}\frac{cos(x+h) – cos(x)}{h}}[/latex]
We can apply the following trig identity:
[latex]\large{cos(x+h) = cos(x)cos(h) – sin(x)sin(h)}[/latex]
Then, rewriting the limit, we have:
[latex]\large{= \lim_{h\to 0}\frac{cos(x)cos(h) – sin(x)sin(h) – cos(x)}{h}}[/latex]
Step 2:
Continuing along, we can factor out cos(x) from two terms as shown:
[latex]\large{= \lim_{h\to 0}\frac{cos(x)[cos(h)-1] – sin(x)sin(h)}{h}}[/latex]
Then, we will apply the limit to each term as shown:
[latex]\large{= cos(x)\lim_{h\to 0}\frac{cos(h)-1}{h}\: – \:sin(x)\lim_{h\to 0}\frac{sin(h)}{h}}[/latex]
Step 3:
We know the value of these limits, so we will substitute them and simplify:
[latex]\large{= cos(x)(0) \: – \: sin(x)(1) = -sin(x)}[/latex]
Thus,
[latex]\frac{d}{dx}cos(x) = -sin(x)[/latex]
Step 1:
Substituting into the definition of the derivative, we have:
[latex]\large{\frac{d}{dx}(\cos(x)) = \lim_{h\to 0}\frac{\cos(x+h) – \cos(x)}{h}}[/latex]
We can apply the following trig identity:
[latex]\small{\cos(x+h) = \cos(x)\cos(h) – \sin(x)\sin(h)}[/latex]
Then, rewriting the limit, we have:
[latex]\large{= \lim_{h\to 0}\frac{\cos(x)\cos(h) – \sin(x)\sin(h) – \cos(x)}{h}}[/latex]
Step 2:
Continuing along, we can factor out cos(x) from two terms as shown:
[latex]\large{= \lim_{h\to 0}\frac{\cos(x)[\cos(h)-1] – \sin(x)\sin(h)}{h}}[/latex]
Then, we will apply the limit to each term as shown:
[latex]= \cos(x)\lim_{h\to 0}\frac{\cos(h)-1}{h}\: – \:\sin(x)\lim_{h\to 0}\frac{\sin(h)}{h}[/latex]
Step 3:
We know the value of these limits, so we will substitute them and simplify:
[latex]= \cos(x)(0) \: – \: \sin(x)(1) = -\sin(x)[/latex]
Thus,
[latex]\frac{d}{dx}\cos(x) = -\sin(x)[/latex]
Derivative of Tangent Function
When discussing the derivative of trig functions, we must also talk about the tangent function. Now if you recall, we know that the tangent function is equal to:
[latex]\large{tan(x) = \frac{sin(x)}{cos(x)}}[/latex]
To find the derivative of tan(x), we can use the quotient rule! Let’s do this as follows:
[latex]\large{\frac{d}{dx}tan(x) = \frac{d}{dx}\frac{sin(x)}{cos(x)}}[/latex]
[latex]\large{= \frac{cos(x)cos(x) – sin(x)(-sin(x))}{cos^2(x)}}[/latex]
[latex]=\large{ \frac{cos^2(x)+sin^2(x)}{cos^2(x)}}[/latex]
Now, if you recall from trig identities, [latex]cos^2(x)+sin^2(x) = 1[/latex]
Hence, we have:
[latex]\large{\frac{d}{dx}tan(x) = \frac{1}{cos^2(x)} = sec^2(x)}[/latex]
When discussing the derivative of trig functions, we must also talk about the tangent function. Now if you recall, we know that the tangent function is equal to:
[latex]\large{tan(x) = \frac{sin(x)}{cos(x)}}[/latex]
To find the derivative of tan(x), we can use the quotient rule! Let’s do this as follows:
[latex]\large{\frac{d}{dx}tan(x) = \frac{d}{dx}\frac{sin(x)}{cos(x)}}[/latex]
[latex]\large{= \frac{cos(x)cos(x) – sin(x)(-sin(x))}{cos^2(x)}}[/latex]
[latex]\large{= \frac{cos^2(x)+sin^2(x)}{cos^2(x)}}[/latex]
Now, if you recall from trig identities,
[latex]\large{cos^2(x)+sin^2(x) = 1}[/latex]
Hence, we have:
[latex]\large{\frac{d}{dx}tan(x) = \frac{1}{cos^2(x)} = sec^2(x)}[/latex]
In a similar manner, we can find the derivative of the remaining three trigonometric functions. The derivative of all the six trig functions are summarized below:
Derivative of trig functions:
[latex]\frac{d}{dx}sin(x) = cos(x)[/latex]
[latex]\frac{d}{dx}cos(x) = -sin(x)[/latex]
[latex]\frac{d}{dx}tan(x) = sec^2(x)[/latex]
[latex]\frac{d}{dx}cot(x) = -csc^2(x)[/latex]
[latex]\frac{d}{dx}sec(x) = sec(x)tan(x)[/latex]
[latex]\frac{d}{dx}csc(x) = -csc(x)cot(x)[/latex]
Example
Find the derivative of the following function:
[latex]\Large{h(x) = \frac{csc(x)}{3x^3}}[/latex]
Step 1:
We need to identify the the function given is a quotient of two other functions, hence, we can use the quotient rule to find the derivative. We can label these two functions as:
[latex]\Large{f(x) = csc(x)}[/latex] and [latex]\large{g(x) = 3x^3}[/latex].
To use the quotient rule, we also need [latex]f'(x)[/latex] and [latex]g'(x)[/latex]. We can easily determine these to be:
[latex]\Large{f'(x) = -csc(x)cot(x)}[/latex] and [latex]\Large{g'(x) = 9x^2}[/latex]
Step 2:
Now, applying the quotient rule:
[latex]\Large{\frac{d}{dx}[\frac{csc(x)}{3x^3}] = \frac{(3x^3)(-csc(x)cot(x)) – (csc(x))(9x^2)}{(3x^3)^2}}[/latex]
Step 3:
Now we can simplify as follows:
[latex]\Large{= \frac{-3x^3csc(x)cot(x) – 9x^2csc(x)}{9x^6}}[/latex]
Factoring out the common term from the numerator, we have:
[latex]\Large{ = \frac{-3x^2csc(x)[xcot(x)+3]}{9x^6}}[/latex]
Finally, our answer is:
[latex]\Large{= \frac{-csc(x)[xcot(x)+3]}{3x^4}}[/latex]
Example
Find the derivative of the following function:
[latex]\Large{h(x) = \frac{csc(x)}{3x^3}}[/latex]
Step 1:
We need to identify the the function given is a quotient of two other functions, hence, we can use the quotient rule to find the derivative. We can label these two functions as:
[latex]\large{f(x) = csc(x)}[/latex] and [latex]\large{g(x) = 3x^3}[/latex].
To use the quotient rule, we also need [latex]f'(x)[/latex] and [latex]g'(x)[/latex]. We can easily determine these to be:
[latex]\large{f'(x) = -csc(x)cot(x)}[/latex] and [latex]\large{g'(x) = 9x^2}[/latex]
Step 2:
Now, applying the quotient rule:
[latex]\large{\frac{d}{dx}[\frac{csc(x)}{3x^3}] = \frac{(3x^3)(-csc(x)cot(x)) – (csc(x))(9x^2)}{(3x^3)^2}}[/latex]
Step 3:
Now we can simplify as follows:
[latex]\large{= \frac{-3x^3csc(x)cot(x) – 9x^2csc(x)}{9x^6}}[/latex]
Factoring out the common term from the numerator, we have:
[latex]\large{ = \frac{-3x^2csc(x)[xcot(x)+3]}{9x^6}}[/latex]
Finally, our answer is:
[latex]\large{= \frac{-csc(x)[xcot(x)+3]}{3x^4}}[/latex]
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