In this article, you will learn what is the derivative of sin x as well as prove the derivative of sin x by quotient rule, first principal rule, and chain rule.
I have already discussed the derivative of cos x, tan x, and sec x in a previous article you can check from here.
So, without wasting time let's get started.
What is sin x?
Sin x is a trigonometric function that is reciprocal of cosec x.
Derivative of sin x
The derivative of sin x is equal to cos x.
We can prove the derivative of sin x in three ways first by using the quotient rule and second by using the first principle rule and the last chain rule.
Derivative of sin x Proof by Quotient Rule
The formula of the quotient rule is,
dy/dx = {v (du/dx) - u (dv/dx)}/v²
Where,
dy/dx = derivative of y with respect to x
v = variable v
du/dx = derivative of u with respect to x
u = variable u
dv/dx = derivative of v with respect to x
v = variable v
Let us,
y = sin x
As we know,
sin x = 1/cosec x
So,
We can write as,
y = 1/cosec x.
Where,
u = 1
v = cosec x
Now putting these values on the quotient rule formula, we will get
dy/dx = [cosec x d/dx(1) - 1 d/dx(cosec x)] / (cosec x)²
Since,
d/dx (cosec x) = -cosec x.cot x
d/dx (1) = 0
So,
dy/dx = {0 - (-cosec x.cot x )}/ (cosec x)²
dy/dx = (cosec x.cot x )/(cosec x)²
dy/dx = (cot x)/(cosec x)
Since,
cot x = cos x/sin x
cosec x = 1/sin x
So,
dy/dx = (cos x/sin x)/(1/sin x)
dy/dx = cos x
d/dx (sin x) = cos x
Thus, we proved the derivative of sin x will be equal to cos x using the quotient rule method.
Derivative of sin x Proof by First Principle Rule
According to the first principle rule, the derivative limit of a function can be determined by computing the formula:
For a differentiable function y = f (x)
We define its derivative w.r.t x as :
dy/dx = f ' (x) = limₕ→₀ [f(x+h) - f(x)]/h
f'(x) = limₕ→₀ [f(x+h) - f(x)]/h
This limit is used to represent the instantaneous rate of change of the function f(x).
Let,
f (x) = sin x
So,
f(x + h) = sin (x + h)
Putting these values on the above first principle rules equation.
f' (x) = limₕ→₀ [sin (x + h) - sin x]/h
As we know,
sin C - sin D = 2 cos{(C + D)/2}.sin{(C - D)/2}
So,
f'(x) = limₕ→₀ [{2cos{(x + h + x)/2}.sin{(x + h - x)/2}] / h
= limₕ→₀ [2 cos{(2x + h)/2}.sin (h/2)]/h
= limₕ→₀ [cos{(2x + h)/2] · limₕ→₀ {sin (h/2)}/(h/2)]
As h →0, and (h/2)→0
So,
f'(x) = limₕ→₀[cos{(2x + h + x)/2} ·lim₍ₕ/₂₎→₀ {sin (h/2)}/ (h/2)]
Now, using limit formulas,
lim ₓ→₀(sin x/x) = 1
So,
f'(x) = [cos{(2x + 0)/2}·(1) = cos (2x/2) = cos x
f'(x) = cos x
Thus, we proved the derivative of sin x will be equal to cos x using the first principle rule method.
Derivative of sin x Proof by Chain Rule
Let us,
y = sin x
As we know,
sin x = cos (π/2 - x)
So,
y = cos (π/2 - x)
By using the chain rule,
The formula of the chain rule is,
dy/dx = (dy/du) × (du/dx)
Where,
dy/dx = derivative of y with respect to x
dy/du = derivative of y with respect to u
du/dx = derivative of u with respect to x
After putting these values we can find,
dy/dx = d/dx {cos (π/2 - x)}
Since,
d/dx(cos x) = -sin x
Hence,
dy/dx = -sin(π/2 - x) · d/dx (π/2 - x)
dy/dx = - sin(π/2 - x).(-1)
dy/dx = sin(π/2 - x)
Since,
sin(π/2 - x) = cos x
Hence,
d/dx (sin x) = cos x
Thus, we proved the derivative of sin x will be equal to the cos x using the chain rule method.
So friends here I discussed all aspects related to the derivative of sin x.
I hope you enjoy this topic If you have any doubts then you can ask me through comments or direct mail. I will definitely reply to you.
Thank You.
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