What is the derivative of the inverse tan(y/x)? - Socratic
2015年5月24日 · The derivative would be 1/sqrt(x^2+y^2) (dy/dx -y/x) If u is tan^-1(y/x) then tan u =y/x. Differentiating w.r.t. x, sec^2u (du)/dx= 1/x^2 (xdy/dx -y) (du)/dx= cos^2 u [1/x^2(x dy/dx -y)] = x/sqrt(x^2+y^2) 1/x^2 (xdy/dx -y) =1/sqrt(x^2+y^2) (dy/dx -y/x)
What is the derivative of inverse tangent of 2x? | Socratic
2015年4月14日 · What is the derivative of inverse tangent of 2x? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 2 Answers
How do you take the derivative of #tan^-1(x^2)#? - Socratic
2015年8月26日 · So, if you have a function #y = tan^(-1)(x^2)#, then you know that you can write . #tan(y) = x^2# Differentiate both sides with respect to #x# to get . #d/(dy)(tany) * (dy)/dx = d/dx(x^2)# #sec^2y * (dy)/dx = 2x# This is equivalent to saying that #(dy)/dx = (2x)/sec^2y# Remember that you have . #color(blue)(sec^2x = 1 + tan^2x)# which means ...
Differentiating Inverse Trigonometric Functions - Calculus - Socratic
The derivatives of inverse trigonometric functions can be computed by using implicit differentiation followed by substitution.
How do you take the derivative of #Tan^-1 (y/x)#? - Socratic
2015年6月7日 · I'm assuming you are thinking of this as being a function of two independent variables x and y: z=tan^{-1}(y/x).
How do you differentiate #y=tan^-1(3x)#? - Socratic
2016年8月12日 · for #d/dx (tan^-1(3x))# you can remember that . #d/(du) ( tan^(-1) u )= 1/(1+u^2)# and that, where #u = u(v)#, via the chain rule: #d/(dv) ( tan^(-1) u )= 1/(1+u^2(u))* (du)/(dv)# or you can switch the function over by saying that . #tan y = 3x# and then differentiating implicitly, so that . #sec^2 y \ y' = 3# BTW you are still using the chain ...
What is the derivative of #f(x)=tan^-1(x)# - Socratic
2015年7月1日 · How do you find the derivative of #y=arcsin(1/x)#? See all questions in Differentiating Inverse Trigonometric Functions Impact of this question
What is the second derivative of inverse tangent? | Socratic
2018年3月3日 · d^2/(dx^2) arctanx= -(2x)/(1+x^2)^2 Let: y = arctanx so that: x=tany differentiate this last equality with respect to x: 1= sec^2y dy/dx Now using the trigonometric inequality: sec^2y = 1+tan^2y we have: 1 = (1+tan^2y)dy/dx 1 =(1+x^2)dy/dx that is: dy/dx =1/(1+x^2) Differentiate again using the chain rule: (d^2y)/(dx^2) = d/dx1/(1+x^2) (d^2y)/(dx^2) = -1/(1+x^2)^2 d/dx …
How do you take the derivative of tan^-1 2x? - Socratic
2015年8月4日 · Use the derivative of tan^-1 and the chain rule. The derivative of tan^-1x is 1/(1+x^2) (for "why", see note below) So, applying the chain rule, we get: d/dx(tan^-1u) = 1/(1+u^2)*(du)/dx In this question u = 2x, so we get: d/dx(tan^-1 2x) = 1/(1+(2x)^2)*d/dx(2x) = 2/(1+4x^2) Note If y = tan^-1x, then tany = x Differentiating implicitly gets us: sec^2y dy/dx = …
How do you find the derivative of the function #y= tan^-1 (3x)
2015年4月14日 · To simplify this, at this point, you could draw a right triangle and label one of the non-right angles #\tan^{-1}(3x)#. Since the tangent of the angle is the length of the opposite side divided by the length of the adjacent side, you could label the opposite side with " #3x# " and the adjacent side with "1".