
What does sec x equal in terms of sin, cos, and/or tan?
2015年9月15日 · sec alpha=1/cos alpha This comes straight from the definition. Secans is defined as inverse of cosine.
Find the Derivative of sec x using first principle? - Socratic
2018年3月7日 · Define the function: # f(x)=secx # Using the limit definition of the derivative, we have: # f'(x) = lim_(h rarr 0) (f(x+h)-f(x))/h #
How do you rrite the trigonometric function sec x in terms
2015年6月2日 · We know that sec x = 1/cos x (Reciprocal identity) We also know that cosx = sin(pi/2-x) = sqrt(1-sin^2x) (Cofunction identity and Pythagorean identity.)
Derivatives of y=sec(x), y=cot(x), y= csc(x) - Calculus - Socratic
The derivatives of \sec(x), \cot(x), and \csc(x) can be calculated by using the quotient rule of differentiation together with the identities \sec(x)=\frac{1}{\cos(x ...
What is the integral of #sec^3(x)#? - Socratic
2014年12月22日 · #I=int sec^3x dx# by Integration by Pats with: #u= secx# and #dv=sec^2x dx# #=> du=secx tanx dx# and #v=tanx#, #=secxtanx-int sec x tan^2x dx# by #tan^2x=sec^2x-1# #=secxtanx-int (sec^3x-secx) dx# since #int sec^3xdx=I#, #=secxtanx-I+int sec x dx# by adding #I# and #int sec x dx=ln|secx+tanx|+C_1# #=>2I=secxtanx+ln|secx+tanx|+C_1# by dividing by 2,
What is Derivatives of #y=sec(x)# - Socratic
2014年9月7日 · #d/dx sec(x)=sec(x)tan(x)# 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)# and #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))# (using the quotient rule)
Fundamental Identities - Trigonometry - Socratic
What does sec x equal in terms of sin, cos, and/or tan? What is the purpose of the cotangent, secant, and ...
How do you simplify (sec(x))^2−1? - Socratic
2015年9月7日 · Using the Pythagorean identity: tan^2x = sec^2x - 1 This is an application of the Pythagorean identities, namely: 1 + tan^2x = sec^2x This can be derived from the standard Pythagorean identity by dividing everything by cos^2x, like so: cos^2x + sin^2x = 1 cos^2x/cos^2x + sin^2x/cos^2x = 1/cos^2x 1 + tan^2x = sec^2x From this identity, we can rearrange the terms to arrive at the answer to your ...
What is the derivative of #sec^-1(x)#? - Socratic
2014年12月11日 · Let y=sec^{-1}x. by rewriting in terms of secant, => sec y=x by differentiating with respect to x, => sec y tan y cdot y'=1 by dividing by sec y tan y, => y' = 1/{sec y tan y} since sec y =x and tan x = sqrt{sec^2 y -1}=sqrt{x^2-1} => y'=1/{x sqrt{x^2-1}} Hence, d/dx(sec^{-1}x)=1/{x sqrt{x^2-1}} I hope that this was helpful.
How do you find a vertical asymptote for y = sec(x)? - Socratic
2014年9月25日 · The vertical asymptotes of y=secx are x={(2n+1)pi}/2, where n is any integer, which look like this (in red). Let us look at some details. y=secx=1/{cosx} In order to have a vertical asymptote, the (one-sided) limit has to go to either infty or -infty, which happens when the denominator becomes zero th