Question

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

Answer 1

Hint: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.