# Find \[\dfrac{d}{{dx}}\cos {(1 - {x^2})^2} = \]

1) \[ - 2x(1 - {x^2})\sin {(1 - {x^2})^2}\]

2) \[ - 4x(1 - {x^2})\sin {(1 - {x^2})^2}\]

3) \[4x(1 - {x^2})\sin {(1 - {x^2})^2}\]

4) None of these

Answer

Verified

105k+ views

**Hint:**Derivative is nothing but a displacement or rate of change of any function. If f(x) is any function with an independent variable x then its derivative can be written as \[y = f'(x)\] where y is the dependent variable (dependent on x). We have many standard derivative formulas to find the derivative that contains trigonometric functions using those formulas we can easily solve this problem.

**Formula:**Some formulas that we need to know:

\[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]

\[\dfrac{d}{{dx}}\cos (x) = - \sin (x)\]

**Complete step by step answer:**

Let us name the given function \[y = \cos {(1 - {x^2})^2}\] as. We aim to find the derivative of the given function \[y\].

Now let’s find the derivative of the function that is given.

\[y = \cos {(1 - {x^2})^2}\]

Before differentiating the function let us substitute \[{(1 - {x^2})^2} = u\] and let us substitute \[z = 1 - {x^2}\]. Thus, the function becomes \[u = {z^2}\].

Then on differentiating u concerning x

\[\dfrac{{du}}{{dx}} = \dfrac{{du}}{{dz}}.\dfrac{{dz}}{{dx}}\] (Since the term \[z\] is a function)

\[\dfrac{{du}}{{dx}} = \dfrac{d}{{dz}}\left( {{z^2}} \right).\dfrac{{dz}}{{dx}}\]

\[\dfrac{{du}}{{dx}} = 2z.\dfrac{{dz}}{{dx}}\]

Now re-substitute the value \[z = 1 - {x^2}\]

\[\dfrac{{du}}{{dx}} = 2(1 - {x^2}).\dfrac{d}{{dx}}(1 - {x^2})\]

\[\dfrac{{du}}{{dx}} = 2(1 - {x^2})(0 - 2x) = - 4x(1 - {x^2})\]

Thus, the function will become \[y = \cos (u)\] where \[{(1 - {x^2})^2} = u\].

Now we will differentiate this function concerning u.

\[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}}.\dfrac{{du}}{{dx}}\] (Since the term \[u\] is a function)

\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\cos (u)\dfrac{{du}}{{du}}\]

\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{du}}\cos (u)\dfrac{{du}}{{dx}}\]

We already found the value of \[\dfrac{{du}}{{dx}} = - 4x(1 - {x^2})\] lets substitute it.

\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{du}}\cos (u)[ - 4x(1 - {x^2})]\]

Now by using the formula \[\dfrac{d}{{dx}}\cos (x) = - \sin (x)\] we get

\[ \Rightarrow \dfrac{{dy}}{{dx}} = [ - \sin (u)][ - 4x(1 - {x^2})]\]

On simplifying this we get

\[ \Rightarrow \dfrac{{dy}}{{dx}} = 4x(1 - {x^2})\sin (u)\]

Now let’s resubstitute the value of u in the above expression.

\[ \Rightarrow \dfrac{{dy}}{{dx}} = 4x(1 - {x^2})\sin {(1 - {x^2})^2}\]

We will stop here since we can’t simplify it further.

Thus, we got the derivative of the given function \[y = \cos {(1 - {x^2})^2}\] as \[\dfrac{{dy}}{{dx}} = 4x(1 - {x^2})\sin {(1 - {x^2})^2}\].

Now let us see the options, option (1) \[ - 2x(1 - {x^2})\sin {(1 - {x^2})^2}\] is not the correct option since we got that \[\dfrac{{dy}}{{dx}} = 4x(1 - {x^2})\sin {(1 - {x^2})^2}\] from our calculation.

Option (2) \[ - 4x(1 - {x^2})\sin {(1 - {x^2})^2}\] is not the correct answer since we got that \[\dfrac{{dy}}{{dx}} = 4x(1 - {x^2})\sin {(1 - {x^2})^2}\] from our calculation.

Option (3) \[4x(1 - {x^2})\sin {(1 - {x^2})^2}\] is the correct answer since we got the same answer in our calculation.

Option (4) None of these is an incorrect answer as we got option (3) as the correct answer.

**Hence, Option (3) \[4x(1 - {x^2})\sin {(1 - {x^2})^2}\] is the correct option.**

**Note:**

As we know that there are many standard formulas for the derivatives of trigonometric functions, using those formulas we can easily find the derivatives of trigonometric functions. If the given function looks complicated, we can use the substitution method to make the given function simpler. After simplification, we need to substitute it to get the original variable.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?