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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)
The answer is y'=-1/ (1+x^2) We start by using implicit differentiation: y=cot^ (-1)x. cot y=x. -csc^2y (dy)/ (dx)=1. (dy)/ (dx)=-1/ (csc^2y) (dy)/ (dx)=-1/ (1+cot^2y) using trig identity: 1+cot^2 theta=csc^2 theta. (dy)/ (dx)=-1/ (1+x^2) using line 2: cot y = x. The trick for this derivative is to use an identity that allows you to substitute ...
Jun 7, 2015. I'm assuming you are thinking of this as being a function of two independent variables x and y: z = tan−1(y x). The answers are ∂z ∂x = − y x2 +y2 and ∂z ∂y = x x2 + y2. Both of these facts can be derived with the Chain Rule, the Power Rule, and the fact that y x = yx−1 as follows:
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 inverse tangent of 2x? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 2 Answers
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 = 1," " so dy/dx = 1/sec^2y From ...
I think he originally intended to do this: dy dx = 1 sec2y. sec2y = 1 + tan2y. tan2y = x → sec2y = 1 +x2. ⇒ dy dx = 1 1 + x2. Answer link. I seem to recall my professor forgetting how to deriving this. This is what I showed him: y = arctanx tany = x sec^2y (dy)/ (dx) = 1 (dy)/ (dx) = 1/ (sec^2y) Since tany = x/1 and sqrt (1^2 + x^2) = sqrt ...
Remember that you have. sec2x = 1 + tan2x. which means that you get. dy dx = 2x 1 + tan2y. Finally, replace tan2y with x2 to get. dy dx = 2x 1 +x4. Answer link. y^' = (2x)/ (1 + x^4) You can differentiate a function y = tan^ (-1) (x^2) by using implicit differentiation. So, if you have a function y = tan^ (-1) (x^2), then you know that you can ...
Answer link. 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 ...
Explanation: u = tan−1(y x) This problem needs a slight prerequisite of chain rule, and quotient rule. = x2 y − xdy dx x2 + y2. Thus, d dx (tan−1(y x)) = x2 y − x dy dx x2 +y2. It really depends upon what you are doing, and which independent variables matter to you. Let z(x,y) = tan−1(y x) ⇒ tanz = y x .