Der of cscx
WebA trigonometric identity relating cscx and sinx is given by cscx = 1 sinx Use of the quotient rule of differentiation to find the derivative of cscx; hence. d dxcscx = d dx( 1 sinx) = ( d dx1)sinx − 1( d dxsinx) sin2x. The derivative … WebSince the derivative of −cot(x) - cot ( x) is csc2 (x) csc 2 ( x), the integral of csc2(x) csc 2 ( x) is −cot(x) - cot ( x). −cot(x)+ C - cot ( x) + C The answer is the antiderivative of the function f (x) = csc2(x) f ( x) = csc 2 ( x). F (x) = F ( x) = −cot(x)+C - cot ( x) + C
Der of cscx
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WebFree Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step WebWhen you have sec x = (cos x)^-1 or cosec x = (sin x)^-1, you have it in the form f (g (x)) where f (x) = x^-1 and g (x) = cos x or sin x. Below is the working for how to derive the derivatives of sec x using this: d/dx (sec x) = d/dx ( (cosx)^-1) = -1 * (cos x)^-2 * d/dx (cos … cscx. = 1/sinx As you can see, if secx= 1/cosx, then sec²x=(1/cosx)² = 1/cos²x, …
WebNov 23, 2024 · d d x ( csc x) = − csc x cot x where sin x ≠ 0 . Proof From the definition of the cosecant function: csc x = 1 sin x From Derivative of Sine Function : d d x ( sin x) = cos x Then: This is valid only when sin x ≠ 0 . Also see Derivative of Sine Function Derivative of Cosine Function Derivative of Tangent Function Derivative of Cotangent Function WebcscX = 1 / sinX sinX = 1 / cscX secX = 1 / cosX cosX = 1 / secX tanX = 1 / cotX cotX = 1 / tanX tanX = sinX / cosX cotX = cosX / sinX Pythagorean Identities sin 2 X + cos 2 X = 1 1 + tan 2 X = sec 2 X 1 + cot 2 X = csc 2 …
WebStudy with Quizlet and memorize flashcards containing terms like der. of tan x, der. of secx, der. of cotx and more. Webd d x ( csc x) = − csc x cot x The derivative or differentiation of cosecant function with respect to a variable is equal to the negative the product of cosecant and cotangent functions. This derivative rule is read as the …
WebDec 17, 2016 · Rewrite cscx in terms of sinx and use the quotient rule. quotient rule y = u v ⇒ dy dx = vu' −uv' v2. y = cscx = 1 sinx. u = 1 ⇒ u' = 0. v = sinx ⇒ v' = cosx. dy dx = (sinx × 0) −(1 ×cosx) (sinx)2. dy dx = 0 −cosx (sinx)2. dy dx = − cosx sinxsinx = − cosx sinx × 1 sinx. dy dx = −cotxcscx.
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