Just do the product rule, followed by the chain rule (on only the #ln(u)#).Recall that #d/(dx)[cosx] = -sinx#, and that #d/(dx)[ln(u(x))] = 1/(u(x))(du(x))/(dx) = 1 ...

Jul 30, 2018 · We know that , #color(red)((1)(a-b)(a^2+ab+b^2)=a^3-b^3# #color(blue)((2)tan(A-B)=(tanA-tanB)/(1+tanAtanB)# #color(brown)((3)lim_(theta to0)(tantheta)/(theta)=1#

Mar 29, 2018 · #tanx/(1-cotx)+cotx/(1-tanx)#, #=(sinx/cosx)/(1-cosx/sinx)+(cosx/sinx)/(1-sinx/cosx)#, #={sinx/cosx-:(sinx-cosx)/sinx}+{cosx/sinx-:(cosx-sinx)/cosx}#,

Nov 7, 2015 · f'(tanx)=sec^2x Using the trig identities, recall that tanx can be rewritten as sinx/cosx. Now differentiate using quotient rule: d/dx(f(x)/g(x)) = (g(x)*f'(x) - f(x)*g'(x))/g(x)^2 …

graph{tan x [-10, 10, -5, 5]}. See explanation. graph{tan x [-10, 10, -5, 5]} The number of lines you have to draw depends on the domain provided. If the domain is not specifically provided, you …

Aug 23, 2017 · but as #tanx# is positive, #sinx# and #cosx# would be positive and negative together. Answer link.

Feb 23, 2016 · It is not an identity. It cannot be verified. The left side is not always equal to the right. For example at x=pi/6 the left side is 4 and the right side is (4sqrt3)/3

Nov 5, 2016 · Tangent: y = 2x+pi/2-1 Normal: y = -1/2x-pi/8 -1 The gradient tangent to a curve at any particular point is given by the derivative. If y=tanx then dy/dx=sec^2x When x=-pi/4 => …

Jul 1, 2016 · C is the correct answer. You need to put on a common denominator. This will be sinxcosx. =(sin^2x + cos^2x)/(sinxcosx) Applying the identity sin^2x + cos^2x = 1: …

May 5, 2016 · As #f(x)=tanx*cotx=1#, #(df(x))/dx=0#.. However, let us also try using product rule.. #df(x)/dx=cotxd/dx(tanx)+tanxd/dx(cotx)#