Quotient Rule Example

Q:  Find dy/dx of

y = x / sqrt(x2 + 1)

A:  To find dy/dx (the derivative), we will need to use the quotient rule since we have a function over a function.  See? We are in the form:

y = f / g where f and g are two different functions of x.

In this form, the quotient rule tells us if:

y = f / g
then

y ‘ = (g * f ‘ – f * g ‘ ) / g2

So, we know:

f = x

g = sqrt(x2 + 1)

f ‘ = 1   <— basic derivative

g ‘ = (this takes more work…. First, rewrite “g”):

g = sqrt(x2 + 1) = (x2 + 1)1/2

Now, use a power rule and a chain rule to find g ‘ like so:

g ‘ = 1/2 (x2 + 1) – 1/2 (2x)

[that’s the derivative of the outside * the derivative of the inside]

Clean up g ‘:

g ‘ = x (x2 + 1) – 1/2

So now we have all of the players:  f, f ‘ , g, g ‘:

f = x

g = (x2 + 1)1/2

f ‘ = 1

g ‘ = x (x2 + 1) – 1/2

Now, we said:  y ‘ = (g * f ‘ – f * g ‘ ) / g2, so plug in the pieces then simplify like so:

y ‘ = [ (x2 + 1)1/2(1) – (x)(x (x2 + 1) – 1/2 ) ] / [(x2 + 1)1/2]2

CLEAN UP

y ‘ = [ (x2 + 1)1/2 – x2(x2 + 1) – 1/2  ] / (x2 + 1)

Multiply top and bottom by (x2 + 1)1/2

[ (x2 + 1)1/2(x2 + 1)1/2 (x2 + 1)1/2x2(x2 + 1) – 1/2  ] / (x2 + 1)1/2(x2 + 1) =

[ (x2 + 1)- x2 ] / (x2 + 1)3/2

1 / (x2 + 1)3/2

TADA!

So, if:

y = x / sqrt(x2 + 1)

dy/dx = 1 / (x2 + 1)3/2

Minimum / Maximum, Concavity, Etc!

Q:  Consider the function:  f(x) = (1/3) x3 – x2 – 15x + 1

Find / tell me about:

(a) critical points

(b) the local minimum / maximums [relative extreme]

(c) points of inflection

(d) concavity

A:  OK.  Get out your pen and paper and be writing with me as I calculate and explain.  This is the best way for you to learn.

(a)  The critical points are where the derivative equals 0 or is undefined.  So, take the derivative to start:

f(x) = (1/3) x3 – x2 – 15x + 1

f ‘ (x) = x2 – 2x- 15  <– you should get this!

Now, we need to find where the derivative = 0 (or is undefined — since the derivative is a quadratic, it is never undefined, so we don’t have to worry about that part).

0 = x2 – 2x- 15

Solve by factoring:

0 = (x – 5)(x + 3)

So, the critical points are when x = 5, -3.

(b)  To determine if the critical points are local maximum or minimums, we need to do the second derivative test.

Recap on second derivative test:  Plug a critical point into the second derivative. 

*If the end result is positive, this means the shape is concave up (smiley face) which means we have a minimum.

*If the end result is negative, this means the shape is concave down (frowny face) which means we have a maximum.

So, take the second derivative:

f(x) = (1/3) x3 – x2 – 15x + 1  [original function]

f ‘ (x) = x2 – 2x- 15  [1st derivative]

f ” (x) = 2x – 2  [2nd derivative]

Now, we have critical points of x = 5 and x = -3.  Deal with them one at a time in the second derivative:

f ” (5) = 2(5) – 2 = 10 – 2 = 8 <– this is positive, which means when x = 5 we have a local minimum!

f ” (-3) = 2(-3) – 2 = -6 – 2 = -8 <– this is negative, which means when x = -3 we have a local maximum!

If you wanted to find the (x, y) point of the actual max and min, you would plug the x-values into the original equations to find the corresponding y values:

f(5) = (1/3) (5)3 – (5)2 – 15(5) + 1 = -172/3

So, f(x) has a local minimum at the point (5, -172/3)

f(-3) = (1/3) (-3)3 – (-3)2 – 15(-3) + 1 = 28

So, f(x) has a local maximum at the point (-3, 28)

(c)  The points of inflection are where the concavity of the function changes from concave up to concave down (and vice versa).  Our POI candidates can be found by setting the second derivative = 0 and finding the x candidates:

f ” (x) = 2x – 2

0 = 2x – 2 and solve for x:

x = 1

So, x = 1 is a candidate for a point of inflection (POI) — it is not a guaranteed POI though.

We need to interval test into the second derivative to see what is happening (generally, you would need to test in between every x value.  We only have 1 x value, so we just need to interval test on either side of it):

Plot your x values on a number line.  In our case, the only value is x = 1.  Pick a number on the left side of 1.  Let’s pick an easy number, 0:

Plug 0 into the second derivative:

f ” (0) = 2(0) – 2 = -2 <– since this is negative:  this interval, to the left of x = 1, is concave down

Now, pick a number to the right of x = 1.  Let’s pick 2:

f ” (2) = 2(2) – 2 = 2 <– since this is positive:  this interval, to the right of x = 1, is concave up

Therefore, when x = 1 we do have a point of inflection (since we changed from concave down to concave up).

If you want the (x, y) point, plug the x value into the original equation to find y:

f(1) = (1/3) (1)3 – (1)2 – 15(1) + 1 = -44/3

There is a point of inflection at (1, -44/3).

(d)  Now to analyze concavity, which we have basically already done in part (c)!

To the left of x = 1, we determined the graph to be concave down.

To the right of x = 1, we determined the graph to be concave up.

Concave down on the interval (-infinity, 1)

Concave up on the interval (1, infinity)
We have done it.  That was a lot to do.  Enjoy.

Implicit Differentiation

Q:  Using implicit differentiation, find dy/dx of

y/(x+6y)=x7-9

at the point (1,-8/49)

Answer:

Taking the derivative of the right side of the equation is fairly basic, taking the derivative of the left side, on the other hand, is harder.  The left side involves a quotient rule.  I am going to ignore the right side of the equation for now and just deal with the left:

y/(x+6y)

The quotient rule tells you (in some form / notation or another):

(bottom)*(deriv. of top) – (deriv of bottom)*(top) all divided by (bottom) squared.  [You may have learned this with f’s and g’s and f ‘ and g ‘ — it all says the same thing]

So:

Top = y

Bottom = x + 6y

Derivative of Top = dy/dx = y’ [whatever notation you are using]

Derivative of Bottom = 1 + 6(dy/dx) = 1 + 6y’

So, plug this into the quotient rule to get:

[(x+6y)*y’ – (1+6y’)*y] / (x + 6y)2

That is the derivative of the left side of the equation.  Take the derivative of the whole thing (left and right) and you get:

[(x+6y)*y’ – (1+6y’)*y] / (x + 6y)2 = 7x6

We now have our derivative.  We need to solve for y’ by plugging in the point that was given to us: (1,-8/49)

Yuck.  This is a lot of algebra.  I don’t have time right now to show all the algebra, but when I plug in the point I get the following:

y’ = -55/343

I checked this solution by hand and on a computer, so I am fairly confident — double check it yourself!

Related Rates

Q:  If y3=16x2, find dx/dt when x=4 and dy/dt=-1

Answer:

First thing is to take the derivative of the equation implicitly to get:

(3y2)dy/dt = (32x)dx/dt

The goal is to solve for dx/dt… So we need to know y, x, and dy/dt.  We know x and dy/dt (given in the problem).  Use the original equation to solve for y when x is 4:

y3=16(4)2

which gives:

y = cuberoot(256)

So, now we know all that we need to know to solve for dx/dt:

(3y2)dy/dt = (32x)dx/dt

Plug in each variable:

(3(cuberoot256)2)(-1) = (32*4)dx/dt

Simplify:

-3*32*cuberoot(2) = (32*4)dx/dt

(-3/4)cuberoot(2) = dx/dt

 

Implicit Differentiation

Q: 

A.  Implicitly find dy/dx of exy=8

B.  Now, solve exy=8 for y first, then take the derivative.  Compare your answers to A and B.

Answer:

A.  Remember, implicit differentiation is just the chain rule:  y is a function of x.  So, we need to use the product rule since we are multiplying two functions ex times y:

Also remember, the derivative of x with respect to x is 1 and the derivative of y with respect to x is dy/dx:

Differentiate as follows:

exy + ex(dy/dx) = 0

Now, solve for dy/dx:

ex(dy/dx) = -exy

dy/dx = -exy/ex

dy/dx = -y [final answer]

However, we may want to put the answer in terms of x.  If this is the case, we know from the original problem:

exy=8 which implies:

y = 8/ex

So, dy/dx = -y = -8/ex

B.  Solve for y first, then take the derivative:

We just solved for y above and got:

y = 8/ex

We can rewrite this as:

y = 8*e-x

Now, take the derivative (using the chain rule)

dy/dx = 8*e-x*(-1)

dy/dx = -8*e-x

Which can then be re-written as:

dy/dx = -8/ex

Compare A and B?  Notice they are identical — which is good news.  Whether we solve for y first or differentiate implicitly, we still get the same answer for dy/dx.

Implicit Differentiation Example

Q:  Find y'(3) using implicit differentiation of the equation:

2x^2 + 3x + xy = 3

and y(3) = -8

Answer:

First, take the derivative of the equation (this will involve a product rule on the “xy” term):

The implicit differentiation of 2x^2 + 3x + xy = 3 is:

4x + 3 + xy’ + y = 0

And we know that when x = 3, y = -8.  So, plug that in and the solve for y’:

4(3) + 3 + 3y’ + (-8) = 0

12 + 3 + 3y’ – 8 = 0

7 + 3y’ = 0

3y’ = -7

y’ = -7/3

So, y'(3) = -7/3

 

Derivative and Power Rule

Q:  Find f ‘ (t) if f(t) = sqrt(3)/t^4

(using the power rule)

Answer:

Step 1)  Rewrite the problem to get t^4 out of the denominator:

sqrt(3)/t^4 = sqrt(3)*t^-4

So, now take the derivative of sqrt(3)*t^-4

Remember, “sqrt(3)” is a constant multiplier, so it just stays along for the ride.  The power rule tells you to bring the exponent down as a multiplier and then subtract 1 from the exponent, like so:

f(t) = sqrt(3)*t^-4

f ‘ (t) = sqrt(3)*(-4)*t^(-4-1)

f ‘ (t) = -4*sqrt(3)t^-5  [final answer]

You could rewrite f ‘ (t) to putt back in the denominator like so:

f ‘ (t) = -4*sqrt(3) / t^5

Using the Definition of a Derivative…

Q:  Use the definition of a derivative to take the derivative of f(x) = 6sqrt(x)

Answer:  First, let’s start with the definition of a derivative:

f ‘ (x) = lim (as h goes to 0) of [f(x+h) – f(x)] / h

So, using our given equation:  f(x) = 6sqrt(x), we have:

f(x+h) = 6sqrt(x+h)

f(x) = 6sqrt(x)

So, the derivative is:

Step 1:  The set-up)  f ‘ (x) = limit as h goes to 0 of [6sqrt(x+h) – 6sqrt(x)]/h

Now, we need to do some simplifying before plugging in 0 for h.

The general strategy to use when you see square roots is to multiply both top and bottom of the fraction by the conjugate, which is (6sqrt(x+h) + 6sqrt(x):

Step 2:  Simplifying)  [6sqrt(x+h) – 6sqrt(x)] * (6sqrt(x+h) + 6sqrt(x)  ) / h * (6sqrt(x+h) + 6sqrt(x)  )

Simplify this more and nice things happen (distribute / FOIL the top):

36(x+h) – 36(x) / [h *(6sqrt(x+h) + 6sqrt(x))]

Simplify numerator more:

(36x + 36h – 36x) / [h *(6sqrt(x+h) + 6sqrt(x))]

36h / [h *(6sqrt(x+h) + 6sqrt(x))]

Cancel out “h”

36 / (6sqrt(x+h) + 6sqrt(x))]

Now, take the limit as h goes to 0 (plug in 0 for h):

36 / (6sqrt(x+0) + 6sqrt(x))] =

36 / (6sqrt(x) + 6sqrt(x))

36 / 12 sqrt(x)

Final answer:  3/sqrt(x)

Marginal Cost, Revenue & Profit

Q:  Word problem, here it is:

(a) The marginal cost function for Janes Coffee Co is MC(q) = 40 – 2q, where q is the number of tons of coffee produced.  Fixed costs are $250.  The marginal revenue function is MR(q) = 100 – 4q.  What is the value of the profit function P(q) when q = 20?

(b) The average cost when q=20 is?

(c) The change in revenue if sales increase from 10 to 15 tons is?

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