Differentiation of Exponential and Logarithmic Functions

[gecko5068]

A few exercises I've been having trouble with, any help is appreciated =) No I'm not begging someone to do my homework, I just need help getting started.

1)EDIT:Solved

2)Find an equation for a line that is normal to the graph of y=xe^x and goes through the origin.

3)Find dy/dx of y=(sinx)^x, 0<x<(pi/2) *SOLVED*

4)The amount A (in grams) of radioactive plutonium remaining in a 20-gram sample after t days is given by the forumula: A=20*(1/2)^(t/140), at what rate is the plutonium decaying when t=2 days? (in grams)

5)The graph of y=lnx looks as though it might be approaching a horizontal asymptote. Write an argument based on the graph of y=e^x to esplain why it does not.

 

[galactus]

3)Find dy/dx of y=(sinx)^x, 0<x<(pi/2)

\(\displaystyle \L\\y=(sin(x))^{x}\)

log of both sides:

\(\displaystyle \L\\ln(y)=xln(sin(x))\)

Differentiate:

\(\displaystyle \L\\\frac{y'}{y}=\frac{x}{tan(x)}+ln(sin(x))\)

Solve for y':

\(\displaystyle \L\\y'=y\left[\frac{x}{tan(x)}+ln(sin(x))\right]\)

Remember that \(\displaystyle y=(sin(x))^{x}\)

\(\displaystyle \L\\y'=(sin(x))^{x}\left[\frac{x}{tan(x)}+ln(sin(x))\right]\)

 

[gecko5068]

x/tanx (in the last step) would simplify to xcotx right?

Thanks a ton for your help =)

 

[gecko5068]

Still stuck... =P :shock:

 

[skeeter]

2)Find an equation for a line that is normal to the graph of y=xe^x and goes through the origin.

tangent line has slope e^x(x + 1) ... normal line has slope -1/[e^x(x+1)]

also good to know, y = xe^x passes thru the origin.

4)The amount A (in grams) of radioactive plutonium remaining in a 20-gram sample after t days is given by the forumula: A=20*(1/2)^(t/140), at what rate is the plutonium decaying when t=2 days? (in grams)

if y = ka^(u), where k is a constant, a is a positive exponential base not equal to 1, and u is a function of t, then

dy/dt = ka^(u)*ln(a)*(du/dt)

5)The graph of y=lnx looks as though it might be approaching a horizontal asymptote. Write an argument based on the graph of y=e^x to esplain why it does not.

y = ln(x) is y = e^x reflected over the line y = x.

 

[gecko5068]

Thanks everyone, I understand it now =).