Calc Studying Help!!

[newParadigm]

I'm working through some old HW problems for practice and I am stumped.

The question reads:

Find an equation of the tangent line to the curve at the given point.

The equation given is:

y=squareroot(x)/(x+1)

The Point given is (4,0.4)

Please help. How would I use derivatives/diferentiation to solve this? I know I need to determine the slope of the Tangent line, and use combine that with the Y-Intercept to form a linear equation, but I just don't see it.

 

[dude8604]

A tangent line is just a line at a certain point whose slope is the slope of the function at that point. The slope is the derivative of the function at that point.

find y'(x) first

y'(x) = ((x+1)(1/sqrt(x)) - sqrt(x))/(x+1)^2

y(4) = ((5/2) - 2))/25) = (1/2)/25 = 1/50

to get the line, do pt-slope form:

y-y1 = m(x-x1) where m is the slope and the line goes through (x1,y1)
just plug in the point and slope and then solve it
=>y-0.4 = (1/50)(x-4) => y=x/50 -4/50 +4/10
=> y = x/50 - 4/50 +20/50 = x/50 +16/50 = x/50 + 8/25

 

[FleshLight]

Find derivative. Substitute. Point slope formula.

 

[newParadigm]

Originally posted by: dude8604
A tangent line is just a line at a certain point whose slope is the slope of the function at that point. The slope is the derivative of the function at that point.

find y'(x) first

y'(x) = ((x+1)(1/sqrt(x)) - sqrt(x))/(x+1)^2

y(4) = ((5/2) - 2))/25) = (1/2)/25 = 1/50

to get the line, do pt-slope form:

y-y1 = m(x-x1) where m is the slope and the line goes through (x1,y1)
just plug in the point and slope and then solve it
=>y-0.4 = (1/50)(x-4) => y=x/50 -4/50 +4/10
=> y = x/50 - 4/50 +20/50 = x/50 +16/50 = x/50 + 8/25

Thankyou bundles...that worked (I actually found it out about a minute before you posted this from a friend on AIM, but thanks nonethless) and gave me the correcdt answer (Since I'm studying I chose to do problems that I could check in the answer key in my book).

 

[DrPizza]

Well, if you check back, here's a hint: for the quotient rule, rather than memorize: "the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all over the denominator squared." Try memorizing this instead: "Ho-d-Hi minus Hi-d-Ho all over Ho Ho" Since I switched to that a few years ago, my students have mastered the quotient rule in a fraction of the time.

(Ho equals denominator, Hi equals numerator, in case that wasn't obvious)

I've heard from many former students who decided to take calc I again at college level that they ended up teaching the quotient rule to a large portion of their class through that mneumonic device.