How to find the derivative of (cosec^2x) - Socratic
May 18, 2018 · differentiate using the chain rule. given y = f (g(x)) then. dy dx = f '(g(x)) × g'(x) ← chain rule. here y = csc2x = 1 sin2x = (sinx)−2. ⇒ dy dx = − 2(sinx)−3 × d dx (sinx) ⇒ dy dx = − 2cosx sin3x. ⇒ dy dx = − 2 × cosx sinx × 1 sin2x. ⇒ dy dx = − 2cotxcsc2x.
Integrate 1÷(1 + cosec x ) ? - Socratic
Apr 4, 2018 · Integrate 1÷(1 + cosec x ) ? Calculus. 2 Answers Shwetank Mauria Apr 4, 2018 #int1/(1+cscx)dx=secx-tanx+x ...
Proving Identities - Trigonometry - Socratic
The Trigonometric Identities are equations that are true for Right Angled Triangles. Periodicity of trig functions. Sine, cosine, secant, and cosecant have period 2π while tangent and cotangent have period π. Identities for negative angles. Sine, tangent, cotangent, and cosecant are odd functions while cosine and secant are even functions.
Prove that cot x + sin x / (1 + cos x) = cosec x - Socratic
Feb 25, 2018 · Prove that cot x + sin x / (1 + cos x) = cosec x ? Trigonometry. 3 Answers Narad T. Feb 25, 2018 See the ...
What is the derivative of cosec x - Socratic
Mar 6, 2018 · 1 Answer. differentiate using the chain rule. given y = f (g(x)) then. dy dx = f '(g(x)) × g'(x) ← chain rule. here y = cscx = 1 sinx = (sinx)−1. ⇒ dy dx = − (sinx)−2 × d dx (sinx) ⇒ dy dx = − cosx sin2x. ⇒ dy dx = − cosx sinx × 1 sinx. ⇒ dy dx = − cotxcscx.
What is the derivative of #y=csc^3(x)#? - Socratic
Mar 3, 2018 · Explanation: We can rewrite y = csc3(x) as y = (cscx)3. Using the Power rule and chain rule, we get. y' = 3(cscx)2 ⋅ d dx (cscx) = 3csc2(x) ⋅ −csc(x)cot(x) = − 3csc3(x)cot(x) As d dx csc(x) = −csc(x)cot(x). Here's a proof of this derivative: csc(x) = 1 sin(x) = (sinx)−1. Since d dx sin(x) = cos(x), using the Power Rule and Chain ...
Prove that ? Sec^2 x + cosec^2 x = sec^2 x . cosec^2 x - Socratic
Jun 5, 2018 · Prove that ? Sec^2 x + cosec^2 x = sec^2 x . cosec^2 x. Trigonometry. 1 Answer Abhishek K. Jun 5, 2018 ...
Derivatives of y=sec(x), y=cot(x), y= csc(x) - Calculus - Socratic
You could memorize this, but you can work it out too by knowing some trig properties. The trig properties we will use are: sec (x)= 1/cos (x) sec (x) = 1 cos (x) and sin x/cos x=tan x sin x cos x = tan x. Deriving: d/dx sec (x) = d/dx 1/cos (x) = (cos (x) (0) - 1 (-sin (x)))/ ( cos (x)cos (x)) d d x sec (x) = d d x 1 cos (x) = cos (x) (0) − 1 ...
Show d/dx (cot x)=-(cosec x)^2.? - Socratic
Mar 5, 2018 · Explanation: We want find the derivative of. y = cot(x) = cos(x) sin(x) Use the quotient rule, if y = f g. then dy dx = f ' ⋅ g − f ⋅ g' g2, thus. f = cos(x) ⇒ f ' = −sin(x) g = sin(x) ⇒ g' = cos(x) Thus. dy dx = −sin(x) ⋅ sin(x) − cos(x)cos(x) sin2(x)
How do you prove csc x - sin x = cos x cot x? - Socratic
Jun 7, 2018 · How do you prove #csc x - sin x = cos x cot x#? Trigonometry Trigonometric Identities and Equations Proving Identities. 2 Answers