The derivative of sin x formula is one of the formulas of differentiation. There are specific formulas in differentiation to find the derivatives of different types of functions. All these formulas are basically derived from the limit definition of the derivative (which is called derivative by the first principle). Here also we are going to prove the derivative of sin x to be -cos x using the first principle.

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Let us learn how to do the differentiation of sin x along with a few examples. Also, let us study the graph of sin x and the derivative of sin x.

1.What is the Derivative of Sin x?
2.Derivative of Sin x Proof by First Principle
3.Derivative of Sin x Proof by Chain Rule
4.Derivative of Sin x Proof by Quotient Rule
5.Graph of Sin x and Derivative of Sin x
6.Derivative of the Composite Function Sin(u(x))
7.FAQs on Derivative of Sin x

What is the Derivative of Sin x?


The derivative of sin x with respect to x is cos x. It is represented as d/dx(sin x) = cos x (or) (sin x)' = cos x. i.e., the derivative of sine function of a variable with respect to the same variable is the cosine function of the same variable. i.e.,

d/dy (sin y) = cos yd/dθ (sin θ) = cos θ

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Derivative of Sin x Formula

The derivative of sin x is cos x. We are going to prove this in each of the following methods.

By first principleBy chain ruleBy quotient rule

Derivative of Sin x Proof by First Principle


The limit definition of the derivative (first principle) is used to find the derivative of any function. We are going to use the first principle to find the derivative of sin x as well. For this, let us assume that f(x) = sin x to be the function to be differentiated. Then f(x + h) = sin(x + h). Now, by the first principle, the limit definition of the derivative of a function f(x) is,

f'(x) = limₕ→₀ / h

Substituting f(x) = sin x and f(x + h) = sin(x + h) here,

f'(x) = limₕ→₀ / h

We can evaluate this limit in two methods.

Method 1

By one of the trigonometric formulas, sin C - sin D = 2 cos <(C + D)/2> sin <(C - D)/2>. Applying this,

f'(x) = limₕ→₀ <2 cos<(x + h + x)/2> sin<(x + h - x)/2> > / h

= limₕ→₀ <2 cos<(2x + h)/2> sin (h/2) > / h

= limₕ→₀ · limₕ→₀ / (h/2)>

As h → 0, (h/2) → 0. So

f'(x) = limₕ→₀ · lim₍ₕ/₂₎→₀ / (h/2)>

Using limit formulas, lim ₓ→₀ (sin x/x) = 1. So

f'(x) = · (1) = cos (2x/2) = cos x

Thus, we have proved that the derivative of sin x is cos x.

Method 2

By sum and difference formulas,

sin (A + B) = sin A cos B + cos A sin B

Using this,

f'(x) = limₕ→₀ / h

= limₕ→₀ < - sin x (1- cos h) + cos x sin h> / h

= limₕ→₀ < - sin x (1 - cos h)>/h + limₕ→₀ (cos x sin h)/h

= -sin x limₕ→₀ (1 - cos h)/h + (cos x) limₕ→₀ sin h/h

Using half angle formulas, 1 - cos h = 2 sin2(h/2).

f'(x) = -sin x limₕ→₀ (2 sin2(h/2))/h + (cos x) limₕ→₀ sin h/h

= -sin x + (cos x) (limₕ→₀ sin h/h)

We know that lim ₓ→₀ (sin x/x) = 1.

f'(x) = -sin x (1 · sin(0/2)) + cos x (1)

= -sin x(0) + cos x (From trigonometric table, sin 0 = 0)

= cos x

Hence we have derived that the derivative of sin x is cos x.


Derivative of Sin x Proof by Chain Rule


By chain rule of differentiation, d/dx(f(g(x)) f'(g(x)) · g'(x). So to find the derivative of sin x using the chain rule, we must write it as a composite function. Using one of the trigonometric formulas, we can write sin x as, sin x = cos (π/2 - x). Using this let us find the derivative of y = sin x (or) cos (π/2 - x).

Using chain rule,

y' = - sin(π/2 - x) · d/dx (π/2 - x) (as the derivative of cos x is - sin x)

= - sin(π/2 - x) · (-1)

= sin(π/2 - x)

= cos x (By one of the trigonometric formulas).

Thus, we have derived the formula of derivative of sin x by chain rule.


Derivative of Sin x Proof by Quotient Rule


The quotient rule says d/dx (u/v) = (v u' - u v') / v2. So to find the differentiation of sin x using the quotient rule, we have to write sin x as a fraction. We know that sin is the reciprocal of the cosecant function (csc). i.e., y = sin x = 1/(csc x). Then by chain rule,

y' = / csc2x

= / csc2x (as the derivative of csc x is -csc x cot x>

= (cot x) / (csc x)

= <(cos x)/(sin x)> / <1/sin x>

= cos x

Therefore, the derivative of sin is cos x and is proved by using the quotient rule.


Graph of Sin x and Derivative of Sin x


The following graph shows the graphs of sin x and its derivative (cos x). We know that a function has a maximum/minimum at a point where its derivative is 0. We can observe in the following graph that wherever sin x is maximum/minimum, cos x is zero at all such points. This way, we can prove that the derivative of sin x is cos x graphically.

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Derivative of the Composite Function Sin(u(x))


sin(u(x)) is a composite function and hence it can be written as sin(u(x)) = f(g(x)) where g(x) = u(x) and f(x) = sin x. Then g'(x) = u'(x) and f'(x) = cos x. We know that the derivative of a composite function is found by using the chain rule. By using chain rule,

d/dx (sin(u(x)) = f'(g(x)) · g'(x)

Since f'(x) = cos x and g(x) = u(x), we have f'(g(x)) = cos (u(x)). So

d/dx (sin(u(x)) = cos (u(x)) · u'(x)

Therefore, the derivative of the composite function sin(u(x)) is cos (u(x)) · u'(x).

Important Notes about Derivative of Sin x:

Here are some important points to note from the differentiation of sin x.

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The derivative of sin x with respect to x is cos x.The derivative of sin u with respect to x is, cos u · du/dx.Sin x is maximum at x = π/2, 5π/2, .... and minimum at x = 3π/2, 7π/2, ...At all these points, the derivative of sin x is 0.i.e., at all these points cos x = 0.

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