Not certain about derivative.

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In summary, the homework statement states that if y=\frac{1-2u}{1+u} and u=\sqrt{x^2-7}, find \frac{dy}{dx} at x=4. The equation for y(u) is y(h) = \frac{1-2h(g(f(x)))}{1+h(g(f(x)))} and the chain rule can be used to simplify this equation.
  • #1
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Homework Statement



If [tex]y=\frac{1-2u}{1+u}[/tex] and [tex]u=\sqrt{x^2-7}[/tex], find [tex]\frac{dy}{dx}[/tex] at [tex]x=4[/tex]

Homework Equations



quotient and chain rule

The Attempt at a Solution



so if [tex]f=y[/tex] then [tex]y^{'}=f^{'}[/tex] then [tex]f^{'} = \left( \frac{f}{g}\right)^{'} = \frac{f^{'}g - fg^{'}}{g^2}[/tex]

and if [tex]f=1-2u[/tex] then [tex]f^{'} = -\left( x^2 -7\right)^\frac{-1}{2} 2x[/tex]

and if [tex]g=1+u[/tex] then [tex]g^{'} = x\left( x^2 -7\right)^\frac{-1}{2}[/tex]

and [tex]g^2 = x^2 + 2\sqrt{x^2 -7}-6[/tex]

and now with chain rule:

[tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\left(\frac{-2\left(1+u\right)-\left(1-2u\right)}{\left(1+u\right)^{2}}\right)\left(x\left(x^{2}-7\right)^{\frac{-1}{2}}\right)[/tex]

i think?
u in terms of x like this:

[tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\left(\frac{-2\left(1+\sqrt{x^2-7}\right)-\left(1-2\sqrt{x^2-7}\right)}{\left(1+\sqrt{x^2-7}\right)^{2}}\right)\left(x\left(x^{2}-7\right)^{\frac{-1}{2}}\right)[/tex]
 
Last edited:
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  • #2
Try the chain rule:

dy/dx =dy/du(du/dx)

should be easy as that
 
  • #3
your f ', g ' and g^2 look correct
 
  • #4
You could also simply substitute u(x) back into the equation, though I would personally go with chain rule.
 
  • #5
okay, hopefully i can simplify it somehow...
 
  • #6
linuxux said:
okay, hopefully i can simplify it somehow...

Did you understand what was said? You appear to be completely ignoring it. YOU put the "chain rule" as one of the relevant equations. The chain rule is specifically to do problems like this without having to substitute one function into another. Use the chain rule, not that complicated formula for y(x) you have!
 
  • #7
HallsofIvy said:
Did you understand what was said? You appear to be completely ignoring it. YOU put the "chain rule" as one of the relevant equations. The chain rule is specifically to do problems like this without having to substitute one function into another. Use the chain rule, not that complicated formula for y(x) you have!

Okay, i will try it again with the chain rule, i don't understand how the chain rule can be used since i can't see a function that's "in" another function? where is it? i actually made a mistake when i wrote chain rule, but since you say its easier i want to do it that way, i will search for a suitable example.

later that day...

ah, now i see, you don't substitute for u as i did, i also see the solution will be much cleaner, thanks!
 
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  • #8
All the mathematicians will probably hate this, but just think of d/dx as something that you want to "cancel" out. So the chain rule says that if you have h(g(f(x))), which you can repeat indefinitely, that you can take the derivative of dh/dx by incrementally "canceling" the derivatives, like so dh/dx = dh/dg*dg/df*df/dx.

So as an example, the say you have [tex] y = sin((x+1)^2) [/tex], which has a form y = h(g(f(x))), where h is the trig function sin(stuff), g is the function inside of the trig function raised to some power (stuff)^2, and f is the function x+1. The solution would be
[tex]\frac{d(sinfunction)}{dg} * \frac{d(powerfunction)}{df} * \frac{d(finalfunction)}{dx} [/tex]
which evaluates to cos((x+1)^2)*2(x^50+1)*50x^49.

You have a function in terms of u, which is also a function of x. So with the chain rule the equation
[tex] y(u) = \frac{1-2u}{1+u}[/tex] where [tex]u(x) = \sqrt{x^2-7}[/tex]

has the form of [tex]y(h) = \frac{1-2h(g(f(x)))}{1+h(g(f(x)))}[/tex]
*remember there is a square root function in u's expression for x, which is a function in itself.

so dy/dx = dy(h)/dg(f)*dg(f)/df(x)*df(x)/dx.

Make sense?
 
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  • #9
Mindscrape said:
All the mathematicians will probably hate this, but just think of d/dx as something that you want to "cancel" out. So the chain rule says that if you have h(g(f(x))), which you can repeat indefinitely, that you can take the derivative of dh/dx by incrementally "canceling" the derivatives, like so dh/dx = dh/dg*dg/df*df/dx.

So as an example, the say you have [tex] y = sin((x+1)^2) [/tex], which has a form y = h(g(f(x))), where h is the trig function sin(stuff), g is the function inside of the trig function raised to some power (stuff)^2, and f is the function x+1. The solution would be
[tex]\frac{d(sin_function)}{dg} * \frac{d(power_function)}{df} * \frac{d(final_function}{dx} [/tex]
which evaluates to cos((x+1)^2)*2(x^50+1)*50x^49.

You have a function in terms of u, which is also a function of x. So with the chain rule the equation
[tex] y(u) = \frac{1-2u}{1+u}[/tex] where [tex]u(x) = \sqrt{x^2-7}[/tex]

has the form of [tex]y(g) = \frac{1-2g(f(x))}{1+g(f(x))}[/tex]

so dy/dx = dy(g)/df(x)*df(x)/dx.

Make sense?

makes perfect sense.
 
  • #10
Catch my edit though, I forgot there was a square root function in the function for u(x).
 
  • #11
yeah i see the change there too.
 
  • #12
Yes, you seem to have caught the sqrt, so your new answer is right. I would put u in terms of x now though.
 
  • #13
is that what you mean when you say u in terms of x?
 
  • #14
Yeah, it's nice to have everything in terms of the variable you a taking the derivative with respect to. :)
 

What is a derivative?

A derivative is a mathematical concept that describes the instantaneous rate of change of a function at a specific point. It represents the slope of the tangent line at that point.

What is the purpose of finding a derivative?

The purpose of finding a derivative is to better understand the behavior and characteristics of a function. It allows us to determine the slope, concavity, and extrema of a function, which can be useful in many real-world applications.

How is a derivative calculated?

A derivative can be calculated using various methods, such as the power rule, product rule, quotient rule, and chain rule. It involves taking the limit of the difference quotient as the change in the input variable approaches zero.

What is the difference between a derivative and an antiderivative?

A derivative measures the instantaneous rate of change of a function, while an antiderivative is the inverse operation of a derivative and finds the original function given its derivative. In other words, a derivative is a rate of change, while an antiderivative is a function.

What are some real-world applications of derivatives?

Derivatives have many applications in fields such as physics, economics, engineering, and finance. Some examples include calculating the velocity and acceleration of an object, optimizing production and profit in a business, and modeling the growth of populations and diseases.

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