Question
Question: If \[y=\left\\{ {{\cot }^{-1}}\sqrt{\dfrac{1+x}{1-x}} \right\\}\] for – 1 < x < 1 then \[\dfrac{dy}{...
If y=\left\\{ {{\cot }^{-1}}\sqrt{\dfrac{1+x}{1-x}} \right\\} for – 1 < x < 1 then dxdy is equal to
Solution
Hint: In order to solve this question, we should know about a few inverse trigonometric ratio derivatives like dxd(cot−1x)=x2+1−1. We should also remember that the derivative of x, that is dxdx=2x1,dxd(f(g(x)))=f′(g(x)).g′(x) and dxd(vu)=v2vdxdu−udxdv. By using these properties, we can solve this question.
Complete step-by-step answer:
In this question, we have been given an equation y=\left\\{ {{\cot }^{-1}}\sqrt{\dfrac{1+x}{1-x}} \right\\} for – 1 < x < 1 and we have been asked to find dxdy. To solve this question, we should have the knowledge of a few standard derivatives like dxdx=2x1,dxd(f(g(x)))=f′(g(x)).g′(x) and dxd(vu)=v2vdxdu−udxdv. Now, to solve this, we will first find the derivative of cot−1 and then the derivative of square root and then the derivative of 1−x1+x. So, we can write,
\dfrac{dy}{dx}=\dfrac{d}{dx}\left\\{ {{\cot }^{-1}}\sqrt{\dfrac{1+x}{1-x}} \right\\}
Now, we know that dxd(cot−1x)=x2+1−1. So, for x=1−x1+x, we get,
dxdy=(1−x1+x)2+1−1.dxd(1−x1+x)
Now, we know that (x)2=x. So, for x=1−x1+x, we get,
dxdy=1−x1+x+1−1.dxd(1−x1+x)
And we can further write it as
dxdy=1−x1+x+1−x−1.dxd(1−x1+x)
dxdy=2−1(1−x).dxd1−x1+x
dxdy=(2x−1)dxd1−x1+x
Now, we know that
dxdx=2x1. So, for x=1−x1+x, we get,
dxdy=(2x−1).21−x1+x1dxd(1−x1+x)
And we can further write it as,
dxdy=(2x−1).(211+x1−x)dxd(1−x1+x)
dxdy=41+x(x−1)1−xdxd(1−x1+x)
Now, we know that dxd(vu)=v2vdxdu−udxdv. So, for u = 1 + x and v = 1 – x, we can write dxdu=1 and dxdv=−1. Therefore, we get,
dxdy=41+x(x−1)1−x[(1−x)2(1−x)(1)−(1+x)(−1)]
Now, we will simplify it further to get dxdy. So, we get,
dxdy=41+x(x−1)1−x[(x−1)21−x+1+x]
dxdy=41+x1−x(x−12)
Now, we will write (x – 1) = – (1 – x). So, we will get,
dxdy=21+x−1−x(1−x1)
dxdy=21+x−(1−x)21−1
dxdy=21+x−(1−x)2−1
And we know that x2−1=x1. So for x = 1 – x, we get,
dxdy=2(1+x)(1−x)−1
And we can further write it as,
dxdy=21−x2−1
Hence, we can say that the derivative of y=cot−11−x1+x for -1 < x < 1 is dxdy=21−x2−1.
Note: While solving this question, we need to be very careful because there are chances of calculation mistakes. Also, we need to remember a few derivatives like dxd(cot−1x)=x2+1−1,dxdx=2x1,dxd(f(g(x)))=f′(g(x)).g′(x) and dxd(vu)=v2vdxdu−udxdv. We might make a mistake while writing the derivatives of cot−1x and vu type derivative. So, we have to be very focused and careful.