Derivative of Sin X

The process to determine the derivative of trigonometric functions is termed differentiation. The alternative definition of differentiation is the rate of change with respect to a given variable. For example, the derivative of the trigonometric function sin x is denoted as sin’ (x) = cos x, it is the rate of change of the function sin x at a specific angle x is stated by the cosine of that particular angle. (i.e) The derivative of sin x is cos x. In this article, we are going to learn what is the derivative of sin x, how to derive the derivative of sin x with a complete explanation and many solved examples.

Derivative of sin x Formula

The derivative of sin x is denoted by d/dx (sin x) = cos x. The other way to represent the sine function is (sin x)’ = cos x.

The derivative of sin x can be found using three different methods, such as:

• By using the chain rule
• By using the quotient rule
• By using the first principle.

Now, let us discuss the first principle method to find the derivative of sin x.

Derivative of sin x using the First Principle Method

The derivative of any function can be found using the limit definition of the derivative. (i.e) First principle. So, now we are going to apply the first principle method to find the derivative of sin x as well.

Assume that the function, f(x) = sin x to be differentiated.

So, f(x+h) = sin (x+h)

By using the first principle for the function f(x), f ’(x) is given by:

f’(x) = lim_h→0 [f(x+h) – f(x)]/h

Substitute f(x) = sin x and f(x+h) = sin(x+h) in the formula, we get

f’(x) = lim_h→0 [ sin (x+h) –  sin(x)]/h

Now, by using the sum and difference of angles in trigonometry, sin (A+B) = sin A cos B + cos A sin B, the above limit can be written as follows:

f’(x) = lim_h→0 [sin x cos h + cos x sin h – sin x]/h

f’(x) = lim_h→0 [-sin x(1-cosh) + cos x sin h]/h

f’(x) = {lim_h→0 [ [-sin x(1-cosh)]/h} + {lim_h→0(cos x sin h)/h}

f’(x) = (- sin x) { lim_h→0 [(1-cosh)]/h} + (cos x) {lim_h→0 (sin h)/h}

Now, by using the half-angle formula, 1- cos h = 2 sin² (h/2), the above equation is written as:

f’(x) = (- sin x) { lim_h→0 [(2 sin² (h/2))]/h} + (cos x) {lim_h→0 (sin h)/h}

f’(x) =(-sin x) [lim_h→0 (sin(h/2))/(h/2). lim_h→0sin (h/2)] + (cos x) {lim_h→0 (sin h)/h}

As we know,

lim_x→0(sin x/x) = 1, we get

 f’(x) = – sin x (1. sin (0/2)) + cos x (1)

f’(x) = – sin x(0) + cos x

f’(x) = cos x

Thus, the derivative of sin x is cos x, is derived.

Derivative of Sin x Examples

Example 1:

Find the derivative of sin (x+1), with respect to x, using the first principle.

Solution:

Assume that f(x) = sin (x+ 1).

Now, we have to find the derivative of sin (x+1), using the 1st principle.

f’(x) = lim_h→0 [f(x+h) – f(x)]/h …(1)

Let f(x) = sin (x+1) and f(x+h) = sin (x+h+1)

Now, substitute the values in (1), we get

f’(x) = lim_h→0 [ sin(x+h+1) – sin(x+1)]/h

f’(x) = lim_h→0 cos (x+h+1).1

f’(x) = cos (x+1).

Hence, the derivative of sin (x+1), with respect to x is cos (x+1).

Example 2:

Find the derivative of sin 2x.

Solution:

To find: derivative of sin 2x.

Given: f(x) = sin 2x

By applying the chain rule, f’(x) is given by

(d/dx) sin 2x = cos 2x (d/dx) 2x

We know that (d/dx) (2x) = 2

Therefore, (d/dx) sin 2x = cos 2x. (2)

Hence, the derivative of sin 2x is 2 cos 2x.

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