
If \[y = {\cos ^{ - 1}}(4{x^3} - 3x)\] then \[\dfrac{{dy}}{{dx}} = ?\]
A.\[\dfrac{3}{{\sqrt {1 - {x^2}} }}\]
B. \[\dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }}\]
C. \[\dfrac{4}{{\sqrt {1 - {x^2}} }}\]
D. \[\dfrac{{ - 4}}{{3{x^2} - 1}}\]
Answer
552.9k+ views
Hint: We substitute the value of x as \[\cos \theta \]and then use the trigonometric formula of \[\cos 3\theta \]to convert the equation in the bracket in simpler form. Cancel the inverse of the given function from the function itself and we get an equation in y and x. Use a method of differentiation to differentiate both sides of the equation.
* \[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \]
* Differentiation of\[{x^n}\]with respect to x is given by \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]
Complete step-by-step solution:
We are given the equation \[y = {\cos ^{ - 1}}(4{x^3} - 3x)\]
Substitute the value of \[x = \cos \theta \]in the given equation
(When \[x = \cos \theta \], then \[\theta = {\cos ^{ - 1}}x\])
\[ \Rightarrow y = {\cos ^{ - 1}}(4{\cos ^3}\theta - 3\cos \theta )\]
Since we know the trigonometric formula \[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \]
Substitute the value of \[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \]in RHS of the equation
\[ \Rightarrow y = {\cos ^{ - 1}}(\cos 3\theta )\]
Cancel inverse cosine function by cosine function in right hand side of the equation
\[ \Rightarrow y = 3\theta \]
Substitute the value of \[\theta = {\cos ^{ - 1}}x\]in right hand side of the equation
\[ \Rightarrow y = 3{\cos ^{ - 1}}x\]
Now differentiate both sides of equation with respect to x
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {3{{\cos }^{ - 1}}x} \right)\]
We can take out constant value from differentiation in right hand side of the equation
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 3\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right)\]
We know \[\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}\], substitute this value of differentiation in right hand side of the equation
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 3 \times \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}\]
Multiply the values in numerator in right hand side of the equation
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }}\]
\[\therefore \]The value of \[\dfrac{{dy}}{{dx}}\] is \[\dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }}\]
\[\therefore \]The correct option is B.
Note: Many students make the mistake of differentiating the given equation directly, which creates a long and confusing answer. Keep in mind when differentiating the inverse trigonometric function the angle inside the function should be as simple as possible so we can substitute the value of angle directly in the differentiation. Also, when we clearly see some part of the equation equal to one of the trigonometric identities, we substitute that value of the equation as a trigonometric term so as to convert it into an easier formula.
* \[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \]
* Differentiation of\[{x^n}\]with respect to x is given by \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]
Complete step-by-step solution:
We are given the equation \[y = {\cos ^{ - 1}}(4{x^3} - 3x)\]
Substitute the value of \[x = \cos \theta \]in the given equation
(When \[x = \cos \theta \], then \[\theta = {\cos ^{ - 1}}x\])
\[ \Rightarrow y = {\cos ^{ - 1}}(4{\cos ^3}\theta - 3\cos \theta )\]
Since we know the trigonometric formula \[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \]
Substitute the value of \[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \]in RHS of the equation
\[ \Rightarrow y = {\cos ^{ - 1}}(\cos 3\theta )\]
Cancel inverse cosine function by cosine function in right hand side of the equation
\[ \Rightarrow y = 3\theta \]
Substitute the value of \[\theta = {\cos ^{ - 1}}x\]in right hand side of the equation
\[ \Rightarrow y = 3{\cos ^{ - 1}}x\]
Now differentiate both sides of equation with respect to x
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {3{{\cos }^{ - 1}}x} \right)\]
We can take out constant value from differentiation in right hand side of the equation
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 3\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right)\]
We know \[\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}\], substitute this value of differentiation in right hand side of the equation
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 3 \times \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}\]
Multiply the values in numerator in right hand side of the equation
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }}\]
\[\therefore \]The value of \[\dfrac{{dy}}{{dx}}\] is \[\dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }}\]
\[\therefore \]The correct option is B.
Note: Many students make the mistake of differentiating the given equation directly, which creates a long and confusing answer. Keep in mind when differentiating the inverse trigonometric function the angle inside the function should be as simple as possible so we can substitute the value of angle directly in the differentiation. Also, when we clearly see some part of the equation equal to one of the trigonometric identities, we substitute that value of the equation as a trigonometric term so as to convert it into an easier formula.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

