WebThe derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide these formulas in the following theorem. Theorem 3.9 Derivatives of tan x, cot x, sec x, and csc x The derivatives of the remaining trigonometric functions are as follows: d d x ( tan x) = sec 2 x (3.13) d d x ( cot x) = − csc 2 x (3.14) WebThe differentiation of trigonometric functions can be done using the derivatives of sin x and cos x by applying the quotient rule. The differentiation formulas of the six …
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WebThe six trigonometric functions have the following derivatives: Theorem 2.17 Derivatives of the Trigonometric Functions For all values of x at which the functions below are defined, we have: (a) d dx (sinx)=cosx (b) d dx (cosx)=−sinx (c) d dx (tanx)=sec2 x (d) d dx (secx)=secxtanx (e) d dx (cotx)=−csc2 x (f) d dx (cscx)=−cscxcotx Webabove functions are shown at the end of this lecture to help refresh your memory: Before we calculate the derivatives of these functions, we will calculate two very important … rcw heal act
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WebDerivatives of Tangent, Cotangent, Secant, and Cosecant We can get the derivatives of the other four trig functions by applying the quotient rule to sine and cosine. For instance, d d x ( tan ( x)) = ( sin ( x) cos ( x)) ′ = cos ( x) ( sin ( x)) ′ − sin ( x) ( cos ( x)) ′ cos 2 ( x) = cos 2 ( x) + sin 2 ( x) cos 2 ( x) = 1 cos 2 ( x) = sec 2 ( x). WebDerivative Proofs of Inverse Trigonometric Functions To prove these derivatives, we need to know pythagorean identities for trig functions. Proving arcsin (x) (or sin-1(x)) … WebSep 7, 2024 · Find the derivative of h(x) = sec(4x5 + 2x). Solution Apply the chain rule to h(x) = sec (g(x)) to obtain h ′ (x) = sec(g(x))tan (g(x)) ⋅ g ′ (x). In this problem, g(x) = 4x5 + 2x, so we have g ′ (x) = 20x4 + 2. Therefore, we obtain h ′ (x) = sec(4x5 + 2x)tan(4x5 + 2x)(20x4 + 2) = (20x4 + 2)sec(4x5 + 2x)tan(4x5 + 2x). Exercise 3.6.3 simultaneous equations worksheet easy