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Question

Question: How do you differentiate \(y = {\cos ^{ - 1}}(1 - 2{x^2})\)?...

How do you differentiate y=cos1(12x2)y = {\cos ^{ - 1}}(1 - 2{x^2})?

Explanation

Solution

In the above question, we have a composite expression therefore, there is no direct formula to find its derivative thus, and we will use the chain rule to find the derivative of the given expression.

Complete step-by-step solution:
We have the given expression as:
y=cos1(12x2)\Rightarrow y = {\cos ^{ - 1}}(1 - 2{x^2})
We have to find the derivative of the given expression therefore; it can be written as:
y=ddxcos1(12x2)\Rightarrow y' = \dfrac{d}{{dx}}{\cos ^{ - 1}}(1 - 2{x^2})
Now the expression in the form of a composite derivative therefore, we will use the chain rule which is: F(x)=f(g(x))g(x)F'(x) = f'(g(x))g'(x)
In this case we have g(x)=12x2g(x) = 1 - 2{x^2}.
Now we know that ddxcos1(xa)=1a2x2\dfrac{d}{{dx}}{\cos ^{ - 1}}\left( {\dfrac{x}{a}} \right) = \dfrac{{ - 1}}{{\sqrt {{a^2} - {x^2}} }}
Now in this case we have no denominator therefore, a=1a = 1, on differentiating, we get:
y=11(12x2)2ddx(12x2)\Rightarrow y' = \dfrac{{ - 1}}{{\sqrt {1 - {{(1 - 2{x^2})}^2}} }}\dfrac{d}{{dx}}(1 - 2{x^2})
Now we know that ddxk=0\dfrac{d}{{dx}}k = 0 and ddxxn=nxn1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} therefore, on differentiating, we get:
y=11(12x2)2×(2(2x))\Rightarrow y' = \dfrac{{ - 1}}{{\sqrt {1 - {{(1 - 2{x^2})}^2}} }} \times (2( - 2x))
On simplifying, we get:
y=4x1(12x2)2\Rightarrow y' = \dfrac{{4x}}{{\sqrt {1 - {{(1 - 2{x^2})}^2}} }}
On expanding the square term in the denominator using the formula (ab)2=a22ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}, we get:
y=4x1(14x2+4x4)\Rightarrow y' = \dfrac{{4x}}{{\sqrt {1 - (1 - 4{x^2} + 4{x^4})} }}
On opening the bracket, we get:
y=4x11+4x24x4\Rightarrow y' = \dfrac{{4x}}{{\sqrt {1 - 1 + 4{x^2} - 4{x^4}} }}
Now on simplifying, we get:
y=4x4x24x4\Rightarrow y' = \dfrac{{4x}}{{\sqrt {4{x^2} - 4{x^4}} }}
Now the term 4x24{x^2} is common in both the terms, we can take it out as common and write the expression as:
y=4x4x2(1x2)\Rightarrow y' = \dfrac{{4x}}{{\sqrt {4{x^2}(1 - {x^2})} }}
Now since the 4x2=2x\sqrt {4{x^2}} = 2x, we take the term out of the square root and write it as:
y=4x2x1x2\Rightarrow y' = \dfrac{{4x}}{{2x\sqrt {1 - {x^2}} }}
Now on simplifying, we get:

y=21x2 \Rightarrow y' = \dfrac{2}{{\sqrt {1 - {x^2}} }}, which is the required solution.

Note: It is to be remembered that chain rule is used only when the expression is in the form of a composite function, which means it is in the form of f(g(x))f(g(x)).
It is to be remembered that integration and differentiation are inverse of each other. If the derivative of the term XX is YY, then inversely, the integration of YY will be XX.
We have used the cos1x{\cos ^{ - 1}}x trigonometric function over here, which is used to find the angle from the value of the trigonometric expression cosx\cos x.