Question

Find the derivative: $\dfrac{{{x^2}\cos \dfrac{\pi }{4}}}{{\sin x}}$.

Answer 1

Hint: To solve this question, we will use the basic formula i.e. quotient rule. We know that the quotient rule is a method of finding the derivative of a function means
if $f(x) = \dfrac{u}{v}$ , then the formula of quotient rule says that,
$f'(x) = \dfrac{{u'v - v'u}}{{{v^2}}}$.
We will use the above formula to solve this question.

Complete step by step answer:
Let us assume that
$f(x) = \dfrac{{{x^2}\cos \dfrac{\pi }{4}}}{{\sin x}}$ .
Now we will assume that
$u = {x^2}\cos \dfrac{\pi }{4}$ and,
$v = \sin x$
So by applying this we can write that we have the function of the form i.e.
$f(x) = \dfrac{u}{v}$
Before applying the formula we will first fund the value of $u'$ and $v'$
We have
$u = {x^2}\cos \dfrac{\pi }{4}$
We know that derivative of the form ${x^n}$ is
 $n{x^{n - 1}}$, where $n$ is the exponential power.
So we can write the derivative of ${x^2}$ as
$2{x^{2 - 1}} = 2x$
In the second part we have constant i.e.
 $\cos \dfrac{\pi }{4}$
So we have
$u' = 2x\cos \dfrac{\pi }{4}$
Now we have
$v = \sin x$
We know that the derivative of $\sin x = \cos x$
So it gives us
$v' = \cos x$
And we can write
${v^2} = \sin x \times \sin x$
It gives the value
${v^2} = {\sin ^2}x$
By putting the values in the quotient rule formula we have:
$\dfrac{{2x\cos \dfrac{\pi }{4}(\sin x) - (\cos x)\left( {{x^2}\cos \dfrac{\pi }{4}} \right)}}{{{{\sin }^2}x}}$
By arranging the above expression can also be written as
$\dfrac{{2x\sin x\cos \dfrac{\pi }{4} - \left( {{x^2}\cos x\cos \dfrac{\pi }{4}} \right)}}{{{{\sin }^2}x}}$
We can take the common factor out and we have
$\dfrac{{x\cos \dfrac{\pi }{4} \left( {2\sin x - x\cos x} \right)}}{{{{\sin }^2}x}}$
Hence the required answer is $\dfrac{{x\cos \dfrac{\pi }{4} \left( {2\sin x - x\cos x} \right)}}{{{{\sin }^2}x}}$.

Note:
We should always remember the rules of differentiation such as the product rule. We know that if the function $f(x)$ is the product of any two functions, i.e.
$f(x) = u(x) \times v(x)$ , then we can say that the derivative of
$f(x) = u'(x) \times v(x) + u(x) \times v'(x)$ .