Quotient Rule

Sully

Am I right in saying the below function is solved using the Quotient Rule?

V(t) = 3/(t+1)^2

Find slop of function at t=1 and find the equation of the tangent at t=1.

So I assume the function then becomes

v=3/(1+1)^2
=3/(2)^2
=3/4

Probably wrong on that part tho.

LeixlipRed

Lot's of issues there. The quotient rule is a method for differentiating a function that itself is one function divided by another.

What you're asked to do is differentiate the function and sub in t=1. What you have done is just sub t=1 into the function. You haven't differentiated.

Delphi91

Sully wrote: »

Am I right in saying the below function is solved using the Quotient Rule?

V(t) = 3/(t+1)^2

Find slope of function at t=1 and find the equation of the tangent at t=1.

So I assume the function then becomes

v=3/(1+1)^2
=3/(2)^2
=3/4

Probably wrong on that part tho.

Yes, you are - you've found the value of V at t=1, not the slope.

The slope of the function is got by calculating dV/dt. You could use the quotient rule for it or else just rewrite the equation as:

[latex]\displaystyle V(t) = 3(t+1)^{-2}[/latex]

Differentiate this and then substitute t=1 to get a value for the actual slope when t=1.

As for finding the equation, using the formula [latex]\displaystyle y - y_1 = m(x - x_1)[/latex], or in your case [latex]\displaystyle V - V_1 = m(t - t_1)[/latex] where m is the slope you just calculated and [latex]\displaystyle (V_1, t_1)[/latex] are the coordinates of the point where the tangent touches the curve (you've already worked this out above!)

Sully

Haven't done this in ages, need to refresh my memory. My maths is appalling to say the least!

I cant recall how to apply the rule to the (t + 1)^2 part. Should I be solving that some other way first, before I can apply the method?

Sully

Why can you re-write the equation like that?

LeixlipRed

http://www.purplemath.com/modules/exponent2.htm

TheBody

Maybe you should check out a few videos on youtube such at this on the quotient rule.

http://www.youtube.com/user/MathTV?feature=chclk#p/search/1/gq4n6xTaVzI