How to Find Derivative of Sin 2x
Here we are going to find the derivative of Sin 2x proof by three methods – First Principle, Chain Rule, Product Rule.
Derivative of Sin 2x Proof by First Principle
To differentiate Sin 2x by First Principal , lets assume f(x)=sin 2x.
Thus consequently using the first principle, we get
⇒ f(x) = lim h->0 [f(x+h)-f(x)]/h
Thus, we can find the derivative as,
⇒f(x) = lim h->0 [sin 2(x+h) – sin 2(x)]/h
⇒f(x) = lim h->0 [sin 2x cos 2h + cos 2x sin 2h – sin 2x]/h , since [sin(A+B) = sin A cos B + cos A sin B]
⇒f(x) = lim h->0 [- sin 2x(1 – cos 2h) + cos 2x sin 2h]/h
Using the half angle formula, 1- cos 2h = 2 sin2 (h),
⇒f(x) = (-sin 2x) { lim [(2 sin2 (h))]/h} + (cos 2x) {lim (sin 2h)/2h}
⇒f(x) = (-sin 2x) [lim (sin(h))/(h). lim sin (h)] + (cos 2x) {lim (sin 2h)/2h}
⇒f(x) = 0 + cos2x (2), since [limx->0 sin 2x/2x = 2]
⇒f(x) = 2 cos 2x
Thus, the required derivative of sin 2x is 2 cos 2x.
Derivative of Sin 2x Proof by Chain Rule
The chain rule computes the derivative of a function combination of two or more functions.
We know the chain rule is,
d/dx(f(g(x))) = d/d(g(x)) [f(g(x)). d/dx (g(x))]
Applying the derivative we get,
⇒ d/dx(sin(2x))
⇒ d/d(2x)(sin(2x)).d/dx(2x)
⇒ cos(2x).2
⇒ 2cos(2x)
Thus using chain rule sin 2x = 2 cox 2x.
Derivative of Sin 2x Proof by Product Rule
Using product rule we easily find the derivative of sin 2x as,
Using the sin double angle formula, Sin2x = 2 sin x cos x
Let u = 2 sin x, and v = cos x. Thus,
⇒ u’ = 2 cos x
⇒ v’ = – sin x
According to product rule:
f(x) = uv’ + vu’
⇒ (2sinx)(–sin x) + (cos x)(2cosx)
⇒ 2(cos2x–sin2x)
⇒ 2cos2x
Thus, using product rule proved that sin 2x = 2 cos 2x.
nth Derivative of Sin 2x
The nth derivative of sin 2x is obtained by differentiating sin 2x n times. To determine nth derivative, we first need first derivative, then second derivative and so on.
Leibniz Formula expresses the derivative of the product of two functions of the nth order. Assuming that the derivatives of the functions u (x) and v (x) are up to nth order
⇒ (uv)(n)=∑ni=0 (ni)u(n−i)v(i)
Thus, the nth derivative of sin 2x is:
⇒ (2sinxcosx)(n) = 2 ∑ni=0 (ni) [sinx]n−i[cosx]i
Here, (ni) indicates the number of i -combinations of elements.
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Derivative of Sin 2x
Derivative of sin 2x is 2cos 2x. Sin 2x is a trigonometric function in which the angle of sin is represented as twice an angle. The trigonometric expansion of sin 2x is 2sinxcosx. The derivative of sin 2x is the rate of change in the function sin 2x to the independent variable x.
In this article, we will learn what is derivative of sin 2x is and how to differentiate sin 2x using various methods in calculus.