How do you solve for x when using the chain rule and functions with variables?

[cogs24]

y = (4-x) ^ 5x

Just wondering what i should do with this
Thanx

 

[Jayboy]

I guess that depends what the question is. Are you asking for the differential equation to which this is the solution?

Take logs and differentiate for first order ODE?

 

[cogs24]

yes, i have to differentiate it

sorry i didnt specifiy what had to be done, i presumed you knew it was differentiation.

 

[Jayboy]

If it's just finding the derivative then this is probably more a question for the calculus thread.

Anyhoo, i'll do the first part. Take logs to get.

ln(y)= 5x ln(4-x)

Now differentiate (implicitly on the LHS) and rearrange to get your answer for dy/dx. Have a go and let me know if/where you get stuck.

 

[cogs24]

ok, i understand your instructions, its just a matter of doing the right things now
This is what i got as an answer, unfortunately we arent supplied with answers in this exercise.

following on from your step, this is what i did

1/y * dy/dx = 5 * (1/4-x)
dy/dx = (5/4-x)y
dy/dx = (5/4-x)(4-x)^5x

 

[dextercioby]

Pay attention with the differentiaition of the RHS.It's a product.I'm sure one of the 2 terms will contain the natural logarithm.

Daniel.

 

[Jayboy]

Think with my poor Tex you missed the x after the 5 on the RHS. Take that into account and you're there.

 

[arildno]

y = (4-x) ^ 5x

Just wondering what i should do with this
Thanx

Here's another way:
Let y(x)=f(u(x),v(x)), f(u,v)=u^{v}, u(x)=4-x, v(x)=5x
Then, we have:
\frac{dy}{dx}=\frac{\partial{f}}{\partial{u}}\frac{du}{dx}+\frac{\partial{f}}{\partial{v}}\frac{dv}{dx}

 

[cogs24]

ahh i see. i didnt spot the product rule on the right, i guess practice makes perfect.
Thanx everyone for the input.

 

[abia ubong]

when differentiating functions of a variable say x raised to another function of xthen let's assume they are T^u,where t and u are functions of x,dy/dx =
T^u[du/dx*logt+u(dt/dx)/t]

 

[abia ubong]

when differentiating functions of a variable say x raised to another function of xthen let's assume they are T^u,where t and u are functions of x,dy/dx =
T^u[du/dx*logt+u(dt/dx)/t].now let 4-x=t,and u=5x.
take normal procedures and see if it works