Question

What is the derivative of \[2\cos x\]?

Answer 1

Hint: We are supposed to find the derivative of \[2\cos x\]. So, first we will notice that 2 is a constant and we will take it outside such that we have to compute 2 times the derivative of \[\cos x\]. Now, we know that the derivative of \[\cos x\] is \[-\sin x\], so we will use this result and find the answer.

Complete step by step solution:
Here we are asked to find out the derivative of \[2\cos .x\]. In order to find that, we need to check whether there is any constant term given or not. If there is a constant term then we are not supposed to differentiate the constant term and it should be left as it is. In order to differentiate the trigonometric term, we have to be well known with the derivatives.
Here is the list of derivatives of the common trigonometric ratios:
\[\begin{align}
  & \left( \sin x \right)^\prime =\cos x \\
 & \left( \cos x \right)^\prime =-\sin x \\
 & \left( \tan x \right)^\prime =\dfrac{1}{{{\cos }^{2}}x} \\
 & \left( \cot x \right)^\prime =-\dfrac{1}{{{\sin }^{2}}x} \\
 & \left( \sec \right)^\prime =\tan x\sec x \\
 & \left( \csc x \right)^\prime =-\cot x\csc x \\
\end{align}\]
Now, let us start solving the given question \[2\cos x\].
We can find that there is a constant term and that is 2. We wouldn’t be differentiating the constant term as told earlier. Putting 2 aside, now we will be differentiating the \[\cos x\] term.
\[\dfrac{d}{dx}(2\cos x)=2\dfrac{d}{dx}\cos x\]
As mentioned above, the derivative of \[\cos x\] is \[-\sin x\].
\[\Rightarrow 2\left( -\sin x \right)=-2\sin x\]
\[\therefore \] The derivative of \[2\cos x=-2\sin x\]

Note: While differentiating, it is mandatory to keep an eye upon the signs so that we get the answer accurately. The most common mistake is assuming the derivative of \[cos x\] to be \[sin x\] . So, we must be aware regarding the derivatives of the basic trigonometric ratios to avoid such mistakes. We can also notice the graph of the function.
The graph of \[-2\sin x\] can be plotted in the following way.