The range of f(x)= sin^(-1)x +cos^(-1)x+ tan^(-1) x is. 01:43. Find the range of f(x)= sin^-1x+tan^-1x+cos^-1x. 02:49. Formula Deck; Free PDF Solutions; All in 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 Unsourced material may be challenged and removed. The following is a list of integrals ( antiderivative functions) of trigonometric functions. For antiderivatives involving both exponential and trigonometric functions, see List of integrals of exponential functions. For a complete list of antiderivative functions, see Lists of integrals. Explanation for the correct options: Step 1. Find the intervals in which x lies: We have given cos - 1 x > sin - 1 x, and we know that, π π sin - 1 x + cos - 1 x = π 2. ⇒ π π cos - 1 x = π 2 - sin - 1 x. But π π π 2 - sin - 1 x > sin - 1 x. ⇒ π π π 2 > 2 sin - 1 x. simplify\:\frac{\sin^4(x)-\cos^4(x)}{\sin^2(x)-\cos^2(x)} simplify\:\frac{\sec(x)\sin^2(x)}{1+\sec(x)} simplify\:\sin^2(x)-\cos^2(x)\sin^2(x) simplify\:\tan^4(x)+2\tan^2(x)+1; simplify\:\tan^2(x)\cos^2(x)+\cot^2(x)\sin^2(x) Show More The derivative of inverse secant function with respect to x is written in limit form from the principle definition of the derivative. d d x ( sec − 1 x) = lim Δ x → 0 sec − 1 ( x + Δ x) − sec − 1 x Δ x. In this case, the differential element Δ x can be written simply as h, if we consider Δ x = h. d d x ( sec − 1 x) = lim h → The inverse trigonometric functions are also known as the anti trigonometric functions or arcus functions. The inverse trigonometric functions of sine, cosine, tangent, cosecant, secant, and cotangent are used to find the angle of a triangle from any of the trigonometric functions. It is widely used in many fields like geometry, engineering According to the law of cosines: ( A B) 2 = ( A C) 2 + ( B C) 2 − 2 ( A C) ( B C) cos ( ∠ C) Now we can plug the values and solve: ( A B) 2 = ( 5) 2 + ( 16) 2 − 2 ( 5) ( 16) cos ( 61 ∘) ( A B) 2 = 25 + 256 − 160 cos ( 61 ∘) A B = 281 − 160 cos ( 61 ∘) A B ≈ 14.3. Problem 2.1. OK, we have x multiplied by cos (x), so integration by parts is a good choice. First choose which functions for u and v: u = x. v = cos (x) So now it is in the format ∫u v dx we can proceed: Differentiate u: u' = x' = 1. Integrate v: ∫ v dx = ∫ cos (x) dx = sin (x) (see Integration Rules) Now we can put it together: Simplify and solve: We have to prove that 3cos−1x = cos−1(4x3 −3x) First consider RHS, cos−1(4x3 −3x) Let us take x =cosθ, we know that cos(3θ) = 4cos3θ−3cosθ. cos−1(4x3 −3x) =cos−1(4cos3θ−3cosθ) = cos−1 (cos(3θ))= 3θ. ∵ θ = cos−1x, cos−1(4x3 −3x) = 3cos−1x. ∴ LH S =RH S. 0 ≤ 3cos−1x ≤π. 0 ≤ cos1 x ≤π/3. ∴ W7mky0H.