Question

How do find the derivative of \[y = co{s^2}\left( x \right)\]?

Answer 1

Hint:
We have given a trigonometric function and we have to find its derivative. The angle of the function is. So we do derivative with respect to \[x\]. We apply derivatives on both sides. Since the trigonometric function is not linear. So firstly the power of the function comes in coefficient and power of function comes in coefficient and power of function decreases by one unit. The derivative of the function comes in product we simplify the result and this will be the derivative of the function.

Complete step by step solution:
The given function is \[y - {\cos ^2}x\] we have to find it derivative differentiating both side with respect to
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\cos }^2}x} \right)\]
\[ \Rightarrow 2{\cos ^{2 - 1}}x.\dfrac{d}{{dx}}\left( {\cos x} \right)\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 2\cos x.\dfrac{d}{{dx}}\left( {\cos x} \right)\]
Now derivatives of \[\cos x\] is \[ - \sin x\]
Therefore \[\dfrac{{dy}}{{dx}} - 2\cos x\left( { - \sin x} \right) = - 2\sin x{\text{ cosx}}\]
Since \[2\sin x{\text{ cosx}}\] is equal to \[\sin 2x\]so

\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \sin 2x\]
This is the required result.

Note:
Trigonometric is the branch of mathematics that studies the relationship between side lengths and angles of the triangle. Trigonometry has six trigonometric functions which are \[\sin ,\cos ,\tan ,\cos ec,\sec \]and \[\cot \].Trigonometric functions are the real functions which related an angle of right functions which related an angle of right angle triangles to the ratio of two sides of triangle.
Trigonometric functions are also called circular functions; we can drive lots of trigonometric formulas. Let we have a function \[y = f(x)\] of variable \[x\]. The derivative of the function is the measure of the rate at which the value \[y\] of the function changes with respect to the variable \[x\].