ravennorton25
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I am awarding this a lot of points because 1, it's calculus and I know no one really wants to do that, and 2 I need actual answers for this. Initially I posted some of these questions but for a lesser point value and now I realize that I won't get the most accurate answers for that so more points! Anyway, to get branliest on this you have to at least give me some reasoning/proof for why this is right so I know you didn't just google this.

So this set of questions is on derivatives of trigonometric and logarithmic functions, and two of them are implicit functions.
1. Find the derivative when y = Ln (sinh 2x)
A. 2 cosh 2x, B. 2 coth 2x, C. 2 sech 2x, D. 2 csch 2x

2. Find the derivative when sinh 3y = cos 2x
A. - 2 sin 2x, B. - 2 (sin 2x/sinh 3y), C. -2/3 tan (2x/3y), D. - (2 sin 2x/3 cosh 3y)

3. Find the derivative when y = sin^2 (4x) cos (3x)
A. 8 sin (4x) cos (3x) - 3 sin^2 (4x) sin (3x)
B. 8 cos (4x) cos (3x) - 3 sin^2 (4x) sin (3x)
C. 8 sin (4x) cos (4x) cos (3x) - 3 sin^2 (4x) sin (3x)
D. 8 sin (4x) cos (4x) cos (3x) - 3 sin^2 (4x) sin (3x) cos (3x)

4. Find the derivative if Ln (x + y) = e^x/y (e is raised to the power of x/y by the way, it isn't e raised to the power of x then over y)
the answer options are way too confusing to type up and have them make any sense so I'm not even going to try on this one

Thank you so much!

Respuesta :

LammettHash
1. [tex]y=\ln(\sinh2x)\implies y'=\dfrac{(\sinh2x)'}{\sinh2x}[/tex]

Recall that [tex](\sinh x)'=\cosh x[/tex], which follows from the definition of the hyperbolic functions:

[tex](\sinh x)'=\left(\dfrac{e^x-e^{-x}}2\right)'=\dfrac{e^x+e^{-x}}2=\cosh x[/tex]

so by the chain rule, the derivative reduces to

[tex]y'=\dfrac{2\cosh2x}{\sinh2x}=2\coth2x[/tex]

2. [tex]\sinh3y=\cos2x\implies(\sinh3y)'=(\cos2x)'\implies3\cosh3y\,y'=-2\sin2x[/tex]

The derivative on the left side follows from the same principle as in the first problem. Solving for [tex]y'[/tex], you get

[tex]y'=-\dfrac{2\sin2x}{3\cosh3y}[/tex]

3. [tex]y=\sin^24x\cos3x[/tex]

Product rule:

[tex]y'=(\sin^24x)'\cos3x+\sin^24x(\cos3x)'[/tex]

then power (for the first derivative) and chain rules:

[tex]y'=2\sin4x(4\cos4x)\cos3x-3\sin^24x\sin3x[/tex]
[tex]y'=8\sin4x\cos4x\cos3x-3\sin^24x\sin3x[/tex]

This can be reduced a bit more, but you can stop here since this is one of the answer choices.

4. [tex]\ln(x+y)=e^{x/y}[/tex]

Chain rule for both sides:

[tex](\ln(x+y))'=\left(e^{x/y}\right)'\implies\dfrac{(x+y)'}{x+y}=\left(\dfrac xy\right)'e^{x/y}[/tex]
[tex]\implies\dfrac{1+y'}{x+y}=\left(\dfrac{y-xy'}y^2\right)e^{x/y}[/tex]

I would stop here, but maybe your answer choices are solutions for [tex]y'[/tex] explicitly. If that's the case, solving [tex]y'[/tex] is a purely algebraic exercise.

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