Dec 31, 2017 · Explanation: You may have learned that we can find the inverse of a function by switching the roles of #x# and #y# and then solve for #y#

f(x) = 2x + (2x^3)/3 + x^5/(20) + x^7/(630) + ... \\ \\ \\ \\ x in RR There are several ways to develop a solution, we could start from first principles and form the Taylor series for the hyperbolic functions using their derivatives; we could also use the definitions of the hyperbolic functions in terms of e^x; or we could use the fact that the Taylor Series for these functions are commonly ...

Feb 10, 2018 · How do you find MacLaurin's Formula for f (x) = sinh x and use it to approximate f (1 2) within 0.01?

Mar 8, 2018 · ∫sinh(x) ⋅ cosh(x) ⋅ dx = 1 2 ∫sin2h(x) ⋅ dx = 1 4 cos2h(x) +C

Mar 3, 2016 · The answer above that uses the limit lim x→0 sinx x also is invalid (using the criteria indicated by the note) because this limit cited needs also L'Hôpital's rule to be improved. It is not correct to say that is an important limit and that is why we must know if we can not prove it in the context that is intended for use.

Oct 31, 2015 · 4sinh^3x*coshx By using the power rule and rules for differentiation for hyperbolic trig functions we get d/dxsinh^4x=4sinh^3x*coshx

Sep 17, 2017 · see below If the question was meant to be I=intsinh^2xcoshxdx by inspection the answer would be 1/3sinh^3x+C by substitution u=sinhx=>du=coshxdx I=intu^2cancel (coshu)xx (du)/cancel (coshu) =intu^2du=1/3u^3+C =1/3sinh^3x+C

May 16, 2018 · Given: ∫ x sinh(x) + cosh(x) dx Multiply the integrand by 1 in the form of sinh(x) − cosh(x) sinh(x) − cosh(x): ∫ x sinh(x) + cosh(x) dx = ∫ x sinh(x) +cosh(x) sinh(x) −cosh(x) sinh(x) −cosh(x) dx The denominator is multiplied using the difference of two squares pattern and the numerator is multiplied using the distributive property: ∫ x sinh(x) + cosh(x) dx = ∫ xsinh(x) − ...

Apr 2, 2015 · Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos (x) and y=tan (x)

Apr 12, 2017 · The answer is =x/ (|x|sqrt (x^2+1) We need (coshx)'=sinhx cosh^2x-sinh^2x=1 sinh^2x=cosh^2x-1 Let rewrite the equation y=cosh^-1 (sqrt (x^2+1)) So, coshy=sqrt (x^2+1) Differentiating both sides (coshy)'= (sqrt (x^2+1))' sinhy*dy/dx= (2x)/ (2sqrt (x^2+1))=x/sqrt (x^2+1) dy/dx=1/sinhy*x/sqrt (x^2+1) We calculate sinhy sinh^2y=cosh^2y-1 =x^2+1-1=x^2 sinhy=|x| …