Jun 28, 2018 · Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x)

Aug 8, 2017 · By the chain rule, #"d"/("d"x) tan^2(sin(x)) = 2*tan(sin(x))* "d"/("d"x) (tan(sin(x)))#, #"d"/("d"x) tan^2(sin(x)) = 2tan(sin(x))sec^2(sin(x))*"d"/("d"x)(sin(x))#,

May 18, 2018 · 0 Apply the definition: tan(x) = \\frac{sin(x)}{cos(x)} So, tan(-pi) = \\frac{sin(-pi)}{cos(-pi)} Actually, you may notice that -pi and pi identify the same angle, since you're …

May 2, 2017 · Let y=arcsinx iff x=siny Then using sin^2A+cos^2A -= 1; we have: sin^2y+cos^2y = 1 => x^2+cos^2y=1 :. cos^2y = 1 -x^2 :. sec^2y = 1/(1 -x^2) And, using the trig ...

Feb 29, 2016 · 0, pi/4, pi/2, (3pi)/4, (5pi)/4, (3pi)/2, (7pi)/4 and 2pi Call x/2 = t . Apply the identity: sin 2a = 2sin a.cos a (sin t)/(cos t) - 2sin t.cos t = 0.

Mar 26, 2017 · Recall that #tanx=sinx/cosx# and the Pythagorean identity #sin^2x+cos^2x=1#. From the identity it can be determined that #sin^2x=1-cos^2x#

Jul 21, 2018 · y'=(1+tan^2(sin(x))*cos(x) We are Using the chain rule (f(g(x))'=f'(g(x))*g'(x) and that (tan(x))'=1+tan^2(x) so we get y'=(1+tan^2(sin(x))*cos(x)

Mar 17, 2018 · Let #x=tantheta=>theta=tan^-1x#. #tan^1(x+sqrt(1+x^2))# #=tan^1(tantheta+sqrt(1+tan^2theta))# #=tan^1(tantheta+sqrt(sec^2theta))#

Apr 28, 2018 · sec^2(sinx)cosx Given: d/dx(tan(sinx)). Here, let y=tan(sinx). Use the chain rule, which states that, dy/dx=dy/(du)*(du)/dx Let u=sinx,:.(du)/dx=cosx.

Apr 29, 2015 · There is another way. Call tan x = t. Use the trig identity: cos 2x = (1 - t^2)/(1 + t^2), and sin 2x = 2t/(1 + t^2) cos 2x = (25 - 36)/(25 + 36) = -11/61 sin 2x = 2(-6/5)/(61/25) = …