Question

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

Answer 1

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.