How do you find #(d^2y)/(dx^2)# given #x=tany#?

Answer 1

# AA y in RR, (d^2y)/dx^2=(-2x)/(1+x^2)^2#.

#x=tan y#.
Case : 1 #"Suppose that,"y in (-pi/2,pi/2) rArr y=arc tan x, x in RR#
#:. dy/dx=1/(1+x^2)#.
#:. (d^2y)/dx^2=d/dx(dy/dx)=d/dx(1/(1+x^2))#
Since, #d/dt(1/t)=-1/t^2#, we have, by the Chain Rule,
#(d^2y)/dx^2=-1/((1+x^2)^2)*d/dx(1+x^2)=(-2x)/((1+x^2)^2)#
Case : 2 #"Next, let "y in (pi/2,3pi/2) rArr pi/2<,y<,3pi/2#
#rArr pi/2-pi<,y-pi<,3pi/2-pi#
#rArr -pi/2<,y-pi<,pi/2, &, tan(y-pi)=-tan(pi-y)=-(-tany)=tany=x#.
Thus, in this Case, #x=tan(y-pi), where, (y-pi) in (-pi/2,pi/2)#
#:." by defn. of arc tan, "y-pi=arc tanx, or, y=pi+arc tan x#
#:. (d^2y)/dx^2=(-2x)/(1+x^2)^2#, as in Case : 1
Thus, # AA y in RR, (d^2y)/dx^2=(-2x)/(1+x^2)^2#.
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Answer 2

To differentiate without using the inverse tangent, see below.

#tan y = x#
Differentiate both sides with respect to #x#.
#d/dx(tanx) = d/dx(x)#
#sec^2y dy/dx = 1#
#dy/dx = cos^2y#
Differentiate again w.r.t. #x#
#(d^2y)/dx^2 = 2cos y (-siny dy/dx)#
Now rfeplace #dy/dx#
#(d^2y)/dx^2 = -2 siny cos^3 y#

To see that this is the same as the other answer

#tany = x# #rArr# #cos y = 1/(1+x^2)#
(To see this draw and label a right triangle with angle #y#, opposite side #x# and adjacent side #1#. So the hypotenuse is #1+x^2#.)
(Or use #tany=x# #rArr# #tan^2y = x^2#, so #1+tan^2y = 1+x^2# and #sec^y = 1+x^2# so that #cos y = 1/(1+x^2)#

Now,

#(d^2y)/dx^2 = -2 siny cos^3 y#
# = -2 siny/cosy cos^4y#
# = -2tany (cos^2y)^2#
# = -2 x 1/(1+x^2)^2#
# = (-2x)/(1+x^2)^2#.
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Answer 3

To find (d^2y)/(dx^2) given x = tan y, we'll start by expressing y in terms of x using the inverse tangent function. Then, we'll differentiate twice with respect to x.

Given x = tan y, we can solve for y: y = arctan(x)

Now, we differentiate y with respect to x: (dy/dx) = d(arctan(x))/dx Using the chain rule, this becomes: (dy/dx) = (1 / (1 + x^2)) * dx/dy

To find (dx/dy), we take the reciprocal of the derivative of x with respect to y: (dx/dy) = 1 / (dy/dx)

Now, we differentiate (dy/dx) with respect to x to find (d^2y)/(dx^2): (d^2y)/(dx^2) = d((dy/dx))/dx Using the chain rule again, this becomes: (d^2y)/(dx^2) = d((dy/dx)/(dx/dy))/dx

Substituting (dy/dx) and (dx/dy) into the equation, we get: (d^2y)/(dx^2) = d((1 / (1 + x^2)) * (dy/dx))/dx

Now, we differentiate this expression with respect to x: (d^2y)/(dx^2) = d((1 / (1 + x^2)) * (1 / (dy/dx)))/dx

Solving this derivative will give us the second derivative (d^2y)/(dx^2) in terms of x.

Therefore, the expression for (d^2y)/(dx^2) given x = tan y is obtained by differentiating the expression (1 / (1 + x^2)) * (1 / (dy/dx)) twice with respect to x.

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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.

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