Question

What is the derivative of \[\sin \left( {ax} \right)\]?

Answer 1

Hint: Here, the given question has a trigonometric function. We have to find the derivative or differentiated term of the function. First consider the function \[y\], then differentiate \[y\] with respect to \[x\] . Here we will use the definition of \[\cosh x\] and \[\sinh x\]to find the derivative of \[\cosh x\].

Complete step-by-step solutions:
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.
Now we know the definition of \[\cosh x\] is \[\dfrac{{{e^x} + {e^{ - x}}}}{2}\].
That is
\[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}\].
Now differentiating this with respect to x,
\[\dfrac{d}{{dx}}\left( {\cosh x} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{e^x} + {e^{ - x}}}}{2}} \right)\]
\[\dfrac{d}{{dx}}\left( {\cosh x} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{e^x}}}{2} + \dfrac{{{e^{ - x}}}}{2}} \right)\]
Using linear combination rule we have
\[\dfrac{d}{{dx}}\left( {\cosh x} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{e^x}}}{2}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{{{e^{ - x}}}}{2}} \right)\]
\[\dfrac{d}{{dx}}\left( {\cosh x} \right) = \dfrac{{{e^x}}}{2} - \dfrac{{{e^{ - x}}}}{2}\]
\[\dfrac{d}{{dx}}\left( {\cosh x} \right) = \dfrac{{{e^x} - {e^{ - x}}}}{2}\].
But we know the definition of \[\sinh x\] is \[\dfrac{{{e^x} - {e^{ - x}}}}{2}\]. Then
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cosh x} \right) = \sinh x\]. This is the required answer.

Thus the required answer is \[\dfrac{d}{{dx}}\left( {\cosh x} \right) = \sinh x\]

Note: We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\]. The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero.