In this note, you will learn:
· Differentiation of sin(x), cos(x) and tan(x)
· Differentiation of sinn f(x), cosn f(x), tann f(x) with the application of Chain Rule
Differentiation of sin(x), cos(x) and tan(x)
In the past few articles on differentiation, we have learnt about:
Therefore, today we are going to be learning about differentiation with relations to trigonometric functions. Let’s begin!
As you can see from the diagram above, we will be learning about the differentiation of the sine, cosine and tangent functions. The derivatives are given by:
d/dx [sin(x)] = cos(x)
d/dx [cos(x)] = -sin(x)
d/dx [tan(x)] = sec2(x)
The derivative of a sine function will give you a cosine function as the output, the derivative of a cosine function will give you a negative sine function as the output and the derivative of a tangent function will give you a secant* squared function as the output.
*Note: A secant function is the reciprocal of a cosine function, while a cosecant function is the reciprocal of a sine function.
If you have difficulty memorizing or am afraid that you might mix up both the trigonometric terms, you can just try to remember that for secant, the third letter of the term is a ‘c’, which stands for cosine. Whereas for cosecant, the third letter of the term is a ‘s’, which stands for sine.
Let’s take a look at a few examples down below:
Differentiate each of the following with respect to x, where x is in radians.
a) 6 sin x b) 3x^2 cos x c) y = (tan x)/ (3x + 5)
Solution:
a) d/dx (6 sin x) = 6 cos x
b) d/dx (3x^2 cos x) = (3x^2) x d/dx (cos x) + (cos x) x d/dx (3x^2)
Note: What we are basically doing in the step above is to separate the algebraic term (3x^2) from the trigonometric term (cos x) and then apply the Product Rule which we have learnt previously. Hence, we will end up with:
= (3x^2) (-sin x) + (cos x) (6x)
= 6x cos x – 3x^2 sin x
c) y = (tan x)/ (3x + 5)
dy/dx = [(3x + 5) x d/dx (tan x) – (tan x) x d/dx (3x + 5)]/ (3x + 5)^2
Note: Since the equation involves a fraction, we are hence able to use the Quotient Rule for this question.
= [(3x + 5) sec^2 x – 3(tan x)]/ (3x + 5)^2
Differentiation of sinn f(x), cosn f(x), tann f(x) with the application of Chain Rule
In the previous point above, the steps to find the derivatives of the trigonometric functions are relatively simple because the angle of the trigonometric function is just a simple variable, x. However, what if the angle is no longer x, but has become a function of x?
In such cases, we must then utilize the Chain Rule we have learnt previously to solve the question. Let us consider the function y = sin f(x).
If we let u = f(x) and y = sin u, hence dy/du = cos u and du/dx = f’(x).
If we were to apply the Chain Rule here, where dy/dx = (dy/du) x (du/dx),
Hence, we will end up with:
d/dx [sin f(x)] = f’(x) cos f(x)
Similarly, with the application of the same method above using Chain Rule, we can obtain the derivative for both cos f(x) and tan f(x) as well, which is as shown below:
d/dx [cos f(x)] = -f’(x) sin f(x)
d/dx [tan f(x)] = f’(x) sec^2 f(x)
In a more detailed context, given that f(x) = ax + b, where a and b are constants, then we will obtain the following results by applying the formulae above:
d/dx [sin (ax + b)] = a cos (ax + b)
d/dx [cos (ax + b)] = -a sin (ax + b)
d/dx [tan (ax + b)] = a sec^2 (ax + b)
Let’s take a look at the examples shown below of the above application:
Differentiate each of the following with respect to x, where x is in radians.
a) 3sin [4x + (π/2)]
b) 5x^2 tan 3x
c) y = (sin 3x)/ [cos (π – 3x)]
d) 6 sin^3 4x
Solution:
a) d/dx {3 sin [4x + (π/2)]} = 3 cos [4x + (π/2)] x 4
Note: For differentiation of trigonometric terms involving functions of x, we must first differentiate the trigonometric function (the ‘outside’ of the brackets) before we differentiate the function of x, which is the ‘inside’ of the brackets.
In this case, the ‘outside’ of the brackets will be the cosine function, and the ‘inside’ of the brackets will be 4x + (π/2).
= 12 cos [4x + (π/2)]
b) d/dx (5x^2 tan 3x) = (5x^2) x d/dx (tan 3x) + (tan 3x) x d/dx (5x^2)
= 5x^2 x [3 x (sec^2 3x)] + (tan 3x) (10x)
= 15x^2 sec^2 3x + 10x tan 3x
= 5x (3x sec^2 3x + 2 tan 3x)
c) y = (sin 3x)/ [cos (π – 3x)]
dy/dx = {[cos (π – 3x)] x d/dx (sin 3x) – (sin 3x) x d/dx [cos (π – 3x)]}/ [cos (π – 3x)]^2
= {[cos (π – 3x)] (3 cos 3x) – (sin 3x) [-3 sin (π – 3x)]}/ [cos (π – 3x)]^2
= [(-cos 3x) (3 cos 3x) – (sin 3x) (-3 sin 3x)]/ (-cos 3x)^2
= [(-3 cos^2 3x) – (-3 sin^2 3x)]/ cos^2 3x
= (-3 cos^2 3x + 3 sin^2 3x)/ cos^2 3x
d) d/dx (6 sin^3 4x) = 6 [3 x (sin^2 4x) x (cos 4x) x 4]
= 72 sin^2 4x cos 4x
Try it for yourself!
Differentiate the following equation with respect to x, where x is in radians.
y = 3 cos (πx + 1)^2 + 6 tan^2[(x/2) + (π/3)]
And that’s all for today, students! Math Lobby hopes that after this article, you have a clear understanding on this topic of differentiation with relations to trigonometry, including the application of chain rule for trigonometric terms with functions of x!
If you have any pending questions, please do go on to our Facebook page, Instagram or contact us directly at Math Lobby! We have certified mathematics tutors to aid you in your journey to becoming a better student!
As always: Work hard, stay motivated and we wish all students a successful and enjoyable journey with Math Lobby!
If you want to receive more Secondary Math Tips from us,
Visit and LIKE our Facebook page at www.facebook.com/mathlobbymotivation/
Visit and FOLLOW our Instagram page at https://www.instagram.com/mathlobbymotivation/
Visit our Website at: www.mathlobby.com
Contact us via SMS/WhatsApp/Telegram/Call +65 96322202
*
*
*
#mathlobby #mathlobbymotivation #bestmathtuition #mathtuitionpotongpasir #mathtuitionsg#freemathresources #sgmaths #sgmathtutor #emath #amath #secschool #mathtutorsg #secondarymathtuition #nleveltuition #mathtuition #secondarymath #mathtutors #tuitionsg#singaporemath #sgtuition #sgstudents #sgmath #singaporetuition #mathstuition #math #maths#mathtutor #tuition #sgparents #secondaryschool
Comments