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What is the derivative of inverse tangent of 2x? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 2 Answers
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:
Recall: int{g'(x)}/{g(x)}dx=ln|g(x)|+C (You can verify this by substitution u=g(x).) Now, let us look at the posted antiderivative. By the trig identity tan x={sin x}/{cos x}, int tan x dx=int{sin x}/{cos x}dx by rewriting it a bit further to fit the form above, =-int{-sin x}/{cos x}dx by the formula above, =-ln|cos x|+C or by rln x=lnx^r, =ln|cos x|^{-1}+C=ln|sec x|+C I hope that this was ...
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)
This is a case of knowing the how the derivative of inverse tangent works, and then following the chain rule. If we were looking at y=arctan(x), there's a way to determine the derivative if you've forgotten the formula. First remember that arctan(x) means "inverse tangent of x," sometimes written as tan^(-1)(x). To invert means to switch the x and the y (among other things, but that's the ...
1 Answer. Wataru. Oct 5, 2014. Recall: (arctanx)' = 1 1 + x2. By Chain Rule, y' = 1 1 + (3x)2 ⋅ (3x)' = 3 1 +9x2. I hope that this was helpful. Answer link.
The derivative of y = arctanx is y' = 1 1 +x2. We can derive this by using implicit differentiation. Since inverse tangent is hard to deal with, we rewrite it as. tan(y) = x. By implicitly differentiating with respect to x, sec2(y) ⋅ y' = 1. By solving for y' and using sec2(y) = 1 + tan2(y),
What is the derivative of this function #y=tan^-1(3x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions
Apr 14, 2015. By the Chain Rule (and knowledge of the derivative of inverse tangent function), the answer is: dy dx = 1 1 + (3x)2 ⋅ 3 = 3 1 +9x2. If you didn't happen to remember the derivative of the inverse tangent function, you could derive the answer by differentiation of both sides of the equation tan(tan−1(3x)) = 3x with respect to x ...