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parabol
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Homework Statement
By using cos and sin subs for tan and sec, find the gradient of:
ln(tan2x+secx)
Homework Equations
tanx=sinx/cosx
secx=1/cosx
The Attempt at a Solution
Substituting
[tex]y=ln(\frac{sin2x}{cos2x}+\frac{1}{cosx})[/tex]
Using the chain rule let:
[tex]z= \frac{sin2x}{cos2x}+\frac{1}{cosx}[/tex]
[tex]y=lnz[/tex]
[tex]\frac{dy}{dz}=\frac{1}{z}=z^{-1}=(\frac{sin2x}{cos2x}+\frac{1}{cosx})^{-1}=\frac{cos2x}{sin2x}+cosx[/tex]
looking at z use the quotient rulee for the first part sin2x/cos2x let:
[tex]u=sin2x[/tex] [tex]\frac{du}{dx}=2cos2x[/tex]
[tex]v=cos2x[/tex] [tex]\frac{dv}{dx}=-2sin2x[/tex]
[tex]\frac{dz}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]
[tex]=\frac{(cos2x)(2cos2x)-(sin2x)(-2sin2x)}{(cos^22x)}[/tex]
[tex]=\frac{(2cos^22x)+(2sin^22x)}{(cos^22x)}[/tex]
[tex]=\frac{2(cos^22x)+(sin^22x)}{(cos^22x)}[/tex]
As [tex]sin^2x+cos^2x=1[/tex] therefore:
[tex]=\frac{2}{(cos^22x)}[/tex]
Going back to the now partionally differentated value of z
[tex]y=\frac{1}{cosx}=cos^{-1}x[/tex]
if [tex]y=cos^{-1}x[/tex] then [tex]x=cosx[/tex]
so [tex]\frac{dx}{dy}=-siny[/tex] so [tex]\frac{dy}{dx}=\frac{-1}{sin y}[/tex]
since - [tex]cos^2y+sin^2y=1[/tex] then [tex]sin^2y=1-cos^2y=1-x^2[/tex]
meaning [tex]siny=\sqrt{1-x^2}[/tex] and [tex]\frac{d}{dx}{cos^{-1}}=\frac{-1}{\sqrt{1-x^2}}[/tex]
That menas that[tex]\frac{dz}{dx}=\frac{2}{(cos^22x)}-\frac{1}{\sqrt{1-x^2}}[/tex]
Back to the fist chain rule:
[tex]\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}=(\frac{cos2x}{sin2x}+cosx).(\frac{2}{(cos^22x)}-\frac{1}{\sqrt{1-x^2}})[/tex]
Sorry for the mass workings but I've been struggling with this and have now had a brain failure and can't see if I am right or even how to simplify the final differential.
Is what I have done correct?