Finding the Gradient of ln(tan2x+secx) Using Substitution | Homework Help

[parabol]

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

$$y=ln(\frac{sin2x}{cos2x}+\frac{1}{cosx})$$

Using the chain rule let:

$$z= \frac{sin2x}{cos2x}+\frac{1}{cosx}$$


$$y=lnz$$

$$\frac{dy}{dz}=\frac{1}{z}=z^{-1}=(\frac{sin2x}{cos2x}+\frac{1}{cosx})^{-1}=\frac{cos2x}{sin2x}+cosx$$


looking at z use the quotient rulee for the first part sin2x/cos2x let:

$$u=sin2x$$ $$\frac{du}{dx}=2cos2x$$

$$v=cos2x$$ $$\frac{dv}{dx}=-2sin2x$$

$$\frac{dz}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$

$$=\frac{(cos2x)(2cos2x)-(sin2x)(-2sin2x)}{(cos^22x)}$$

$$=\frac{(2cos^22x)+(2sin^22x)}{(cos^22x)}$$

$$=\frac{2(cos^22x)+(sin^22x)}{(cos^22x)}$$

As $$sin^2x+cos^2x=1$$ therefore:

$$=\frac{2}{(cos^22x)}$$


Going back to the now partionally differentated value of z

$$y=\frac{1}{cosx}=cos^{-1}x$$

if $$y=cos^{-1}x$$ then $$x=cosx$$

so $$\frac{dx}{dy}=-siny$$ so $$\frac{dy}{dx}=\frac{-1}{sin y}$$

since - $$cos^2y+sin^2y=1$$ then $$sin^2y=1-cos^2y=1-x^2$$

meaning $$siny=\sqrt{1-x^2}$$ and $$\frac{d}{dx}{cos^{-1}}=\frac{-1}{\sqrt{1-x^2}}$$

That menas that$$\frac{dz}{dx}=\frac{2}{(cos^22x)}-\frac{1}{\sqrt{1-x^2}}$$


Back to the fist chain rule:

$$\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}=(\frac{cos2x}{sin2x}+cosx).(\frac{2}{(cos^22x)}-\frac{1}{\sqrt{1-x^2}})$$



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?

 

[HallsofIvy]

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

$$y=ln(\frac{sin2x}{cos2x}+\frac{1}{cosx})$$

Using the chain rule let:

$$z= \frac{sin2x}{cos2x}+\frac{1}{cosx}$$


$$y=lnz$$

$$\frac{dy}{dz}=\frac{1}{z}=z^{-1}=(\frac{sin2x}{cos2x}+\frac{1}{cosx})^{-1}=\frac{cos2x}{sin2x}+cosx$$

No! 1/(a+ b) is NOT equal to (1/a) + (1/b).

looking at z use the quotient rulee for the first part sin2x/cos2x let:

$$u=sin2x$$ $$\frac{du}{dx}=2cos2x$$

$$v=cos2x$$ $$\frac{dv}{dx}=-2sin2x$$

$$\frac{dz}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$

$$=\frac{(cos2x)(2cos2x)-(sin2x)(-2sin2x)}{(cos^22x)}$$

$$=\frac{(2cos^22x)+(2sin^22x)}{(cos^22x)}$$

$$=\frac{2(cos^22x)+(sin^22x)}{(cos^22x)}$$

As $$sin^2x+cos^2x=1$$ therefore:

$$=\frac{2}{(cos^22x)}$$


Going back to the now partionally differentated value of z

$$y=\frac{1}{cosx}=cos^{-1}x$$

if $$y=cos^{-1}x$$ then $$x=cosx$$

so $$\frac{dx}{dy}=-siny$$ so $$\frac{dy}{dx}=\frac{-1}{sin y}$$

since - $$cos^2y+sin^2y=1$$ then $$sin^2y=1-cos^2y=1-x^2$$

meaning $$siny=\sqrt{1-x^2}$$ and $$\frac{d}{dx}{cos^{-1}}=\frac{-1}{\sqrt{1-x^2}}$$

That menas that$$\frac{dz}{dx}=\frac{2}{(cos^22x)}-\frac{1}{\sqrt{1-x^2}}$$


Back to the fist chain rule:

$$\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}=(\frac{cos2x}{sin2x}+cosx).(\frac{2}{(cos^22x)}-\frac{1}{\sqrt{1-x^2}})$$



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?

 

[parabol]

No! 1/(a+ b) is NOT equal to (1/a) + (1/b).

Thanks for the input. Have I started off using the method by using the chain rule?

Do I just leave it as:

$$(\frac{sin2x}{cos2x}+\frac{1}{cosx})^{-1}$$

?

 

[Bohrok]

You have another mistake:
$$\frac{1}{cosx} \neq cos^{-1}x = arccosx$$
You can look here on how to differentiate something like 1/f(x) here
http://en.wikipedia.org/wiki/Reciprocal_rule
Look at the last example on the page

One more note: it would have been much easier to differentiate tan2x using u=2x than changing it to sin2x/cos2x, although you did get that part correct.

Thanks for the input. Have I started off using the method by using the chain rule?

Do I just leave it as:

$$(\frac{sin2x}{cos2x}+\frac{1}{cosx})^{-1}$$

?

That, or put it in the denominator as /(tan2x + secx), or find a common denominator and add them, then take the reciprocal to get rid of the ^-1.

 

[parabol]

I'm still not 100% with this question.

I have been re-working on it and have now come up with this answer.

$$\frac{dy}{dx}=\frac{\frac{2(sin2x)^2}{(cos2x)^2}+\frac{sinx}{(cosx)^2}+2}{\frac{sin2x}{cos2x}+\frac{1}{cosx}}$$

Does this look right?

 

[Bohrok]

Where did that last 2 come from in the numerator at the end? If you take that out, you have the correct derivative of ln(tan2x + secx).