Trig Differentiation Help (1 Viewer)

ludac · Apr 10, 2010

f(x) = 1-sinx/cosx. Find f'(x). I used the Quotient rule, but cant get right answer. The right answer is -1/1+sinx . Any help would be appreciated. Here is my attempt: = cosx.cosx-(1-sinx).-sinx/(cosx)^2 = -cos^2x + sinx(1-Sinx)/(cosx)^2 I dunno wat to do after this

Flying_Penis · Apr 10, 2010

What da faak. Just sin/cos = tan
deriv of tan is sec^2 x
Sin^2 + cos^2 = 1 use that.
You should be able to simply that into something useful
Also, remember using sec^2 = 1/cos^2

dishab · Apr 10, 2010

K from carrying on ur answer.
=-cos^2x - sin^2x + sinx all over cos^2x

= -1 + sinx all over cos^2x.
Thats a gud enough answer.

randomnessss · Apr 14, 2010

ludac said:
f(x) = 1-sinx/cosx. Find f'(x). I used the Quotient rule, but cant get right answer. The right answer is -1/1+sinx . Any help would be appreciated. Here is my attempt: = cosx.cosx-(1-sinx).-sinx/(cosx)^2 = -cos^2x + sinx(1-Sinx)/(cosx)^2 I dunno wat to do after this

$\begin{align*}f'(x)&=\frac{cosx.(-cosx)-(-sinx)(1-sinx)}{(cosx)^2} \\&=\frac{-cos^2 x-(-sinx+sin^2 x)}{cos^2 x} \\&=\frac{-cos^2 x+sinx-sin^2 x}{cos^2 x} \\&=-\frac{cos^2 x+sin^2 x-sinx}{cos^2 x} \\&=-\frac{cos^2 x+sin^2 x-sinx}{cos^2 x} \\&=-\frac{1-sinx}{cos^2 x} \\&=-\frac{1-sinx}{1 - sin^2 x} \\&=-\frac{1-sinx}{(1 - sin x)(1+sinx)} \\&=-\frac{1}{(1+sinx)}\end{align*}$

Trig Differentiation Help (1 Viewer)

ludac

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Flying_Penis

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dishab

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randomnessss

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