Question

How do you find the derivative of $cos\left( {\sqrt x } \right)$?

Answer 1

Hint:
In this question, we want to find the derivative of the cosine square root of x. First, consider it as a function $y = \cos \left( {\sqrt x } \right)$. Based on the definition of differentiation, we will solve this question. The formula is $\dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}}$ . Then apply some trigonometry formulas and simplify the question.

Complete step by step solution:
The function given in this question is,
$ \Rightarrow y = \cos \left( {\sqrt x } \right)$
To solve the above function, we will apply the formula of differentiation.
$ \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}}$ ...(1)
Substitute the value of the function in equation (1).
$ \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\cos \left( {\sqrt {x + \Delta x} } \right) - \cos \left( {\sqrt x } \right)}}{{\Delta x}}$
Now, we already know the trigonometry formula $\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$.
Here, the value of ‘A’ is $\sqrt {x + \Delta x} $ and the value of ‘B’ is $\sqrt x $.
Apply this formula and substitute the values of A and B.
$ \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{ - 2\sin \left( {\dfrac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)\sin \left( {\dfrac{{\sqrt {x + \Delta x} - \sqrt x }}{2}} \right)}}{{\Delta x}}$...(2)
Now, we can write,
$ \Rightarrow \left( {\sqrt {x + \Delta x} + \sqrt x } \right)\left( {\sqrt {x + \Delta x} - \sqrt x } \right) = {\left( {\sqrt {x + \Delta x} } \right)^2} - {\left( {\sqrt x } \right)^2}$
We can cancel out the square root and square on the right-hand side.
Therefore,
$ \Rightarrow \left( {\sqrt {x + \Delta x} + \sqrt x } \right)\left( {\sqrt {x + \Delta x} - \sqrt x } \right) = x + \Delta x - x$
Apply the subtraction on the right-hand side. The subtraction of x and x is 0.
Therefore,
$ \Rightarrow \left( {\sqrt {x + \Delta x} + \sqrt x } \right)\left( {\sqrt {x + \Delta x} - \sqrt x } \right) = \Delta x$
Let us put this value in equation (2).
$ \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{ - 2\sin \left( {\dfrac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)\sin \left( {\dfrac{{\sqrt {x + \Delta x} - \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)\left( {\sqrt {x + \Delta x} - \sqrt x } \right)}}$
Let us split the denominator,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{ - 2\sin \left( {\dfrac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \times \dfrac{{\sin \left( {\dfrac{{\sqrt {x + \Delta x} - \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} - \sqrt x } \right)}}$
Divide the numerator and the denominator by 2.
So,
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\sin \left( {\dfrac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \times \dfrac{{\sin \left( {\dfrac{{\sqrt {x + \Delta x} - \sqrt x }}{2}} \right)}}{{\left( {\dfrac{{\sqrt {x + \Delta x} - \sqrt x }}{2}} \right)}}$
Apply the limit to both functions.
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\sin \left( {\dfrac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \times \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\sin \left( {\dfrac{{\sqrt {x + \Delta x} - \sqrt x }}{2}} \right)}}{{\left( {\dfrac{{\sqrt {x + \Delta x} - \sqrt x }}{2}} \right)}}$
Now, let us consider $\left( {\dfrac{{\sqrt {x + \Delta x} - \sqrt x }}{2}} \right) = u$.
Here, as $\Delta x \to 0$ then $u \to 0$.
So,
 $ \Rightarrow \dfrac{{dy}}{{dx}} = - \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\sin \left( {\dfrac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \times \mathop {\lim }\limits_{u \to 0} \dfrac{{\sin \left( u \right)}}{{\left( u \right)}}$
 We know that $\mathop {\lim }\limits_{u \to 0} \dfrac{{\sin \left( u \right)}}{{\left( u \right)}} = 1$
So,
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\sin \left( {\dfrac{{\sqrt {x + \Delta x} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \times 1$
Now, apply the limit to the function.
So,
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\sin \left( {\dfrac{{\sqrt {x + 0} + \sqrt x }}{2}} \right)}}{{\left( {\sqrt {x + 0} + \sqrt x } \right)}}$
That is equal to,
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\sin \left( {\dfrac{{2\sqrt x }}{2}} \right)}}{{\left( {2\sqrt x } \right)}}$
Therefore,
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\sin \left( {\sqrt x } \right)}}{{2\sqrt x }}$

Hence the answer is $ - \dfrac{{\sin \left( {\sqrt x } \right)}}{{2\sqrt x }}$.

Note:
The second method to solve the question is as below.
We want to find the derivative of cosine square root x.
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cos \left( {\sqrt x } \right)} \right)$
Apply the formula $\dfrac{d}{{dx}}\cos y = - \sin y\dfrac{{dy}}{{dx}}$
Put $\sqrt x $ instead of y in the above formula.
So,
 $ \Rightarrow \dfrac{d}{{dx}}\left( {\cos \left( {\sqrt x } \right)} \right) = - \sin \left( {\sqrt x } \right)\dfrac{d}{{dx}}\left( {\sqrt x } \right)$
We can write $\sqrt x $ as ${x^{\dfrac{1}{2}}}$.
So,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cos \left( {\sqrt x } \right)} \right) = - \sin \left( {\sqrt x } \right)\dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right)$
Now, apply the formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ .
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cos \left( {\sqrt x } \right)} \right) = - \sin \left( {\sqrt x } \right)\left( {\dfrac{1}{2}} \right)\left( {{x^{\dfrac{1}{2} - 1}}} \right)$
So,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cos \left( {\sqrt x } \right)} \right) = - \dfrac{1}{2}\sin \left( {\sqrt x } \right)\left( {{x^{\dfrac{{1 - 2}}{2}}}} \right)$
 That is equal to,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cos \left( {\sqrt x } \right)} \right) = - \dfrac{1}{2}\sin \left( {\sqrt x } \right)\left( {{x^{ - \dfrac{1}{2}}}} \right)$
We know that ${x^{ - 1}} = \dfrac{1}{x}$ .
So,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cos \left( {\sqrt x } \right)} \right) = - \dfrac{{\sin \left( {\sqrt x } \right)}}{{2{x^{\dfrac{1}{2}}}}}$
We can write ${x^{\dfrac{1}{2}}}$ as $\sqrt x $.
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cos \left( {\sqrt x } \right)} \right) = - \dfrac{{\sin \left( {\sqrt x } \right)}}{{2\sqrt x }}$