Find the derivative of #y=3tan^-1(x+sqrt(1+x^2))#?

Answer 1

#(dy)/(dx)=3/(2(1+x^2)#

We know that.

#color(violet)((1)cos(pi/2-theta)=sintheta and sin(pi/2-theta)=costheta#
#color(blue)((2) (1+cosalpha)=2cos^2(alpha/2) and sinalpha=2sin(alpha/2)cos(alpha/2)#
#color(brown)((3)cot(pi/2-alpha)=tanalpha#
#color(red)((4)tan^-1(tan alpha)=alpha ,where, alpha in(-pi/2,pi/2)#

Here,

#y=3tan^-1(x+sqrt(1+x^2))#

Substitute, #x=tantheta=>theta=tan^-1x,where,thetain (- pi/2,pi/2)#

#:.y=3tan^-1(tantheta+sqrt(1+tan^2theta))#
#=3tan^-1(tantheta+sectheta)#
#=3tan^-1(sintheta/costheta+1/costheta)#
#=3tan^-1((1+sintheta)/costheta)..to Apply(1)#
#=3tan^-1((1+cos(pi/2-theta))/sin(pi/2-theta))...toApply(2)# #=3tan^-1((2cos^2(pi/4-theta/2))/(2sin(pi/4-theta/2)cos(pi/4- theta/2)))#
#=3tan^-1(cos(pi/4-theta/2)/sin(pi/4-theta/2))#
#=3tan^-1(cot(pi/4-theta/2))#
#=3tan^-1(cot{pi/2-pi/4-theta/2})#
#=3tan^-1(cot(pi/2-(pi/4+theta/2))...toApply(3)#
#=3tan^-1(tan(pi/4+theta/2))..toApply(4)#

Now,

#theta in (-pi/2,pi/2)#
#=>theta/2 in(-pi/4,pi/4)...to #[Dividing by #2#]
#=>pi/4+theta/2 in (-pi/4+pi/4,pi/4+pi/4)...to#[Adding #pi/4#]
#=>pi/4+theta/2 in(0,pi/2)sub(-pi/2, pi/2)#

So,

#y=3(pi/4+theta/2)#
#y=(3pi)/4+3/2*theta,where, theta=tan^-1x#
#=>y=(3pi)/4+3/2tan^-1x#
#=>(dy)/(dx)=0+3/2(1/(1+x^2))#
#i.e. (dy)/(dx)=3/(2(1+x^2)#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the derivative of ( y = 3 \tan^{-1To find the derivative of ( y = 3 \tan^{-1}(The derivativeTo find the derivative of ( y = 3 \tan^{-1}(x +The derivative ofTo find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{The derivative of (To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1The derivative of ( y = 3To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + xThe derivative of ( y = 3\tTo find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^The derivative of ( y = 3\tanTo find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2The derivative of ( y = 3\tan^{-To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2})The derivative of ( y = 3\tan^{-1To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) \The derivative of ( y = 3\tan^{-1}(To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ),The derivative of ( y = 3\tan^{-1}(x+\To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), weThe derivative of ( y = 3\tan^{-1}(x+\sqrtTo find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll useThe derivative of ( y = 3\tan^{-1}(x+\sqrt{1To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use theThe derivative of ( y = 3\tan^{-1}(x+\sqrt{1+xTo find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chainThe derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) \To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) )To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) isTo find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{dThe derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \fracTo find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dxThe derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx}The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \leftThe derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 +To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tanThe derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(uThe derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{1To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(u) \The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{1+x^To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(u) \right]The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{1+x^2To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(u) \right] =The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{1+x^2})To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(u) \right] = \The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{1+x^2})^To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(u) \right] = \fracThe derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{1+x^2})^2}To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(u) \right] = \frac{The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{1+x^2})^2} \To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(u) \right] = \frac{1}{The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{1+x^2})^2} ).To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(u) \right] = \frac{1}{1The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{1+x^2})^2} ).To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(u) \right] = \frac{1}{1 +The derivative of ( y = 3\tan^{-1}(x+\sqrt{1+x^2}) ) is ( \frac{3}{1 + (x+\sqrt{1+x^2})^2} ).To find the derivative of ( y = 3 \tan^{-1}(x + \sqrt{1 + x^2}) ), we'll use the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(u) \right] = \frac{1}{1 + u^2} \cdot \frac{du}{dx} ]

First, let ( u = x + \sqrt{1 + x^2} ). Then, we find ( \frac{du}{dx} ):

[ \frac{du}{dx} = 1 + \frac{1}{2\sqrt{1 + x^2}} \cdot 2x = 1 + \frac{x}{\sqrt{1 + x^2}} ]

Now, applying the chain rule:

[ \frac{d}{dx} \left[ \tan^{-1}(x + \sqrt{1 + x^2}) \right] = \frac{1}{1 + (x + \sqrt{1 + x^2})^2} \cdot \left(1 + \frac{x}{\sqrt{1 + x^2}}\right) ]

[ = \frac{1}{1 + (x^2 + 2x\sqrt{1 + x^2} + 1 + x^2)} \cdot \left(1 + \frac{x}{\sqrt{1 + x^2}}\right) ]

[ = \frac{1}{2x^2 + 2x\sqrt{1 + x^2} + 2} \cdot \left(1 + \frac{x}{\sqrt{1 + x^2}}\right) ]

[ = \frac{1 + \frac{x}{\sqrt{1 + x^2}}}{2x^2 + 2x\sqrt{1 + x^2} + 2} ]

Finally, multiply by 3 to get the derivative of the original function:

[ \frac{dy}{dx} = 3 \cdot \frac{1 + \frac{x}{\sqrt{1 + x^2}}}{2x^2 + 2x\sqrt{1 + x^2} + 2} ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7