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Derivative of Sin?

TopTrigonometric Functions are used to show the relationship between the angles of triangle to the lengths of sides of a triangle.. Let’s talk about the derivative of a function f (u) which is given by:

$\frac{d}{du}$ f(u)
or

f'(u)

This given function is also known as derivative or we can say differentiation with respect to ‘u’. In some of the cases the differentiation of a function f (u) is known as differentiation coefficient of f (u).

Let’s talk about the derivative of sin (x);

The derivative of sin (x) is shown below:

Sin (x) = cos (x);

Now we will see the proof of derivative of sin (x);

First we write the sin (x) in the derivative form:

d / dx Sin (x) = cos (x);

Here we apply the Chain Rule for finding the derivative of sin (x).

Chain rule is used to differentiate compositions of Functions. The chain rule is given by:

D f (g(x)) = f’ (g (x)) g’(x);

Here we use the Limit Definition of the derivative along with the sum of the angles of sin x.

f'(x) = $\lim_{\Delta x \to 0}$ $\frac{sin(x + \Delta x) - sin x}{\Delta x}$

We know that the formula for sin (A + B) = sin A. cos B + cos A. sin B; then put in the given expression.

f'(x) = $\lim_{\Delta x \to 0}$ $\frac{sin x cos \Delta x + cos x sin \Delta x - sin x}{\Delta x}$

On further solving we get:

f'(x) = cos x $\lim_{\Delta x \to 0}$ $\frac{ sin \Delta x}{\Delta x}$ + sin x $\lim_{\Delta x \to 0}$ $\frac{cos \Delta x - 1}{\Delta x}$


Then put the limits in the expression we get:

=

f'(x) = cos x + sin x * 0

On further solving we get:

0 + cos (x);

$\frac{d}{dx}$ Sin x = cos x.