Solveeit Logo

Question

Question: What is the derivative of \(\sin (-2x)\) ?...

What is the derivative of sin(2x)\sin (-2x) ?

Explanation

Solution

First, we need to analyze the given information which is in the trigonometric form.
\bullet We can equate the given expression into some form and then we can differentiate using the derivatives of basic functions and applying the chain rule of differentiation.
\bullet The trigonometric functions are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Formula used:
Chain rule of differentiation ddx(f(g(x))=f1(g(x))×g1(x)\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)

Complete step-by-step solution:
Since from the given that we have sin(2x)\sin (-2x) and we need to find its derivative part. Also, note that 22 is a constant function and xx is the variable to the given sine function.
Thus, using the chain rule of the differentiation, we have
ddx(f(g(x))=f1(g(x))×g1(x)\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)
ddx(sin(2x))=d(sin(2x))d(2x)×d(2x)dx\Rightarrow \dfrac{d}{{dx}}(\sin (-2x)) = \dfrac{{d(\sin (-2x))}}{{d(-2x)}} \times \dfrac{{d(-2x)}}{{dx}}
Since we know the derivative of the sinx\sin x with respect to the variable xx is ddx(sinx)=cosx\dfrac{d}{{dx}}(\sin x) = \cos x
Now we proceed to further steps using the formula given
ddx(sin(2x))=d(sin(2x))d(2x)×d(2x)dx\dfrac{d}{{dx}}(\sin (-2x)) = \dfrac{{d(\sin (-2x))}}{{d(-2x)}} \times \dfrac{{d(-2x)}}{{dx}} d(sin(2x))d(2x)×2dxdx\Rightarrow \dfrac{{d(\sin (-2x))}}{{d(-2x)}} \times -2\dfrac{{dx}}{{dx}}
Since the derivative of the constant function will not change.
Then we get,
ddx(sin(2x))=d(sin(2x))d(2x)×d(2x)dx\dfrac{d}{{dx}}(\sin (-2x)) = \dfrac{{d(\sin (-2x))}}{{d(-2x)}} \times \dfrac{{d(-2x)}}{{dx}} cos(2x)×2\Rightarrow \cos (-2x) \times -2
Hence, we have ddx(sin(2x))=2cos(2x)=2cos(2x)\dfrac{d}{{dx}}(\sin (-2x)) = -2\cos (-2x)= -2\cos(2x) As we know cos(θ)=cos(θ)\cos(-\theta)=\cos(\theta)
Thus, the derivative of the given trigonometric function sin(2x)\sin (-2x) is 2cos(2x)-2\cos (2x).

Note: The concept used in the given problem is the chain rule is the main thing. We must know the derivatives of the basic functions like a sine.
Differentiation is defined as the derivative of independent variable value and can be used to calculate features in an independence variance per unit modification.
In differentiation, the derivative of xx raised to the power is denoted by ddx(xn)=nxn1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} .
Differentiation and integration are inverse processes like a derivative of d(x2)dx=2x\dfrac{{d({x^2})}}{{dx}} = 2x and the integration is 2xdx=2x22x2\int {2xdx = \dfrac{{2{x^2}}}{2}} \Rightarrow {x^2}
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like sincos=tan\dfrac{{\sin }}{{\cos }} = \tan and tan=1cot\tan = \dfrac{1}{{\cot }}.