Question

How do you differentiate $k(x) = - 3\cos x$ ?

Answer 1

In this the function k(x) is given a trigonometric function. We have to find the derivative or differentiated term of the function. First consider function k(x)=y, then differentiate y with respect to x by using a standard differentiation formula of trigonometric ratio. And on further simplification we get the required differentiate value.

Formula used:
In the trigonometry we have standard differentiation formula
the differentiation of cos x is -sin x that is $\dfrac{d}{{dx}}(\cos x) = - \sin x$

Complete step by step answer:
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Consider the given function
$k(x) = - 3\cos x$---------- (1)
Now we have to differentiate this function with respect to x, to get the derivative of the function.First replace the function $k(x) = y$, then equation (1) becomes
$y = - 3\cos x$
Differentiate function y with respect to x
$\dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( { - 3\cos x} \right)$
Where -3 is a constant then RHS becomes
$\dfrac{{dy}}{{dx}} = - 3 \cdot \dfrac{d}{{dx}}\left( {\cos x} \right)$---------- (2)

Using the standard differentiated formula of trigonometric ratio cosine is $\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$, then equation (2) becomes
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = - 3 \cdot \left( { - \sin x} \right)$
As we know the sign convention $- \times - = +$, then RHS becomes
$\dfrac{{dy}}{{dx}} = 3\sin x$---------- (3)
Replace $\dfrac{{dy}}{{dx}} = k'(x)$, $k'(x)$ is a differentiated term of function $k(x)$.
Equation (3) becomes
$\therefore k'\left( x \right) = 3\sin x$

Hence the differentiated term of the given trigonometric function $k(x) = - 3\cos x$ is $3\sin x$.

Note: The student must know about the differentiation formulas for the trigonometry ratios and these differentiation formulas are standard. If the function is a product of two terms and the both terms are the function of $x$ then we use the product rule of differentiation to the function.