2018年2月27日 · tan−1(x + 1 x − 1) + tan−1(x − 1 x) = − tan−1 7 Find x?

2016年5月27日 · Let sin−1x = θ hence x = sinθ For 0 <x <1 we draw a right triangle with hypotenuse equal to 1 and the other side equals to x like the one in the Figure below.

2015年4月13日 · dy dx = 1 1 +x2 Let y = tan−1(x) tany = x Differentiating both sides with respect to 'x' sec2y dy dx = 1 dy dx = 1 sec2y As sec2 = 1 +tan2y So, dy dx = 1 1 + tan2y dy dx = 1 1 +x2

2018年7月30日 · Let y = tan−1x Then, tany = x Implicit differentiation yields sec2y dy dx = 1 dy dx = 1 sec2y But 1 + tan2y = sec2y sec2y = 1 + x2 Finally, dy dx = 1 1 +x2

2018年4月17日 · Given: H = tan−1(y x) Substitute x = u + v and y = u −v: H = tan−1(u −v u +v) Let g = u − v u + v, then use the chain rule: dH dv = d(tan−1(g)) dg dg dv d(tan−1(g)) dg = 1 g2 + 1 …

2018年4月17日 · -pi/2 lim_ (x->oo)tan^-1 (x^2-x^4)=tan^-1 (lim_ (x->oo)x^2-x^4) We can move the limit inside as the arctangent is a continuous function. Evaluating, we obtain tan^-1 (lim_ (x …

2015年9月28日 · Probably the best way to find this is to construct the Taylor series centered at x = 0 for tan−1(x) and use the pattern to find the appropriate coefficient, and then, ultimately, the …

2015年4月17日 · 1. Firstly find the derivative of arctanx... 2. Lastly, use implicit differentiation to find the derivative of y= (arctanx)^ (1/2)

2016年4月14日 · Please see below. d/dx (e^tan (1/x)) = e^tan (1/x) [d/dx (tan (1/x))] = e^tan (1/x) [sec^2 (1/x) d/dx (1/x)] = e^tan (1/x) sec^2 (1/x) [ (-1/x^2)] = - (e^tan (1/x ...

2016年12月24日 · The derivative of the tangent is sec^2, so we have sec^2 (1/x) as the first half of our derivative. This is f' (g (x)) in the above definition. Next, we multiply by the derivative of …