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Question

Question: What is the derivative of \(\dfrac{d}{{dx}}\cos \left( {ax} \right)\)?...

What is the derivative of ddxcos(ax)\dfrac{d}{{dx}}\cos \left( {ax} \right)?

Explanation

Solution

Hint : Here, in the given question, we are given a trigonometric function cos(ax)\cos \left( {ax} \right). We have to find the derivative of the given function with respect to xx. Since we have to differentiate with respect to xx but the function is axax. So, to find the derivative of the given trigonometric function we will apply the chain rule. Here two functions on which chain rule will be applied will be cos(x)\cos \left( x \right) and axax.
Formula used
ddxf(g(x))=f(g(x)).g(x)\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)
In words: differentiate the outside function, and then multiply by the derivative of the inside function.

Complete step-by-step answer :
To solve the given trigonometric function we will apply chain rule. According to chain rule, if we are given two functions ff and gg then we find ddxf(g(x))\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) as,
ddxf(g(x))=f(g(x)).g(x)\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)
Let f(x)=cosxf\left( x \right) = \cos x and g(x)=axg\left( x \right) = ax, then
ddxcos(ax)=dd(ax)cos(ax)×ddx(ax)\Rightarrow \dfrac{d}{{dx}}\cos \left( {ax} \right) = \dfrac{d}{{d\left( {ax} \right)}}\cos \left( {ax} \right) \times \dfrac{d}{{dx}}\left( {ax} \right)
ddxcos(ax)=dd(ax)cos(ax)×adxdx\Rightarrow \dfrac{d}{{dx}}\cos \left( {ax} \right) = \dfrac{d}{{d\left( {ax} \right)}}\cos \left( {ax} \right) \times \dfrac{{adx}}{{dx}}
Since we know that the derivative of cos(x)\cos \left( x \right) with respect to xx is given by ddxcos(x)=sinx\dfrac{d}{{dx}}\cos \left( x \right) = - \sin x and derivative of xx with respect to xx is 11. Therefore, we get
ddxcos(ax)=sin(ax)×a\Rightarrow \dfrac{d}{{dx}}\cos \left( {ax} \right) = - \sin \left( {ax} \right) \times a
ddxcos(ax)=asin(ax)\Rightarrow \dfrac{d}{{dx}}\cos \left( {ax} \right) = - a\sin \left( {ax} \right)
Hence, the derivative of the given trigonometric function ddxcos(ax)\dfrac{d}{{dx}}\cos \left( {ax} \right) is asin(ax) - a\sin \left( {ax} \right).
So, the correct answer is “ asin(ax) - a\sin \left( {ax} \right)”.

Note : Differentiation is a process of finding the rate of change of a function. Here, we have found the rate of change of cos(ax)\cos \left( {ax} \right) with respect to xx. Remember that the derivative of a constant with respect to any variable is zero. Also, the derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function. The given question can also be solved by using the first principle of differentiation. First principle: Suppose ff is a real function and xx is a point in its domain of definition. The derivative of ff at xx is defined by limh0f(x+h)f(x)h\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h} provided this limit exists.