BAM! Mr Tarrou. In this video we're going to be doing three examples of finding derivatives using
the quotient rule down below you'll find time stamps to jump ahead to any one of the three
examples you're also going to find a link to my home page of my youtube channel where you can find
any one of the 700 plus math videos i've done to date for algebra one two count two
and statistics so our first two examples we have here is the derivative of secant of x
and the derivative uh with respect to x of four x over x squared plus six so what are we to do
when we go to take a derivative of basically the division of two different functions a
quotient well the derivative with respect to x of f of x over g of x is going to be equal to
g of x times f prime of x minus f of x times g prime of x over x g of x squared
well that's all great but this looks rather complicated and the difference between working
with the product rule and the quotient rule as we start learning these sort of faster methods
of finding the derivative is that the quotient rule has a subtraction in here now we know that
subtraction is not commutative so it's very very important that we remember the order
of the subtraction in the numerator otherwise we can do everything else
right but get the sign of our final answer incorrect so while this is the formula for
the derivative of a quotient the derivative of two functions being divided together it is
at least in my class this is how i remember the rule low for derivative low d high d for
derivative so low d high the denominator times the derivative of the numerator low d high minus high
d low high meaning the numerator and d low of course now we're representing the the the the
derivative of the denominator so low d high or so let me change the way i'm pointing this low d high
minus high d low all over low low the denominator squared now what we have
here with our first example is the derivative with respect to x of the secant of x and you
might be asking yourself why is this the first example i don't see a division of two functions
well up to up till now all you may know is the derivative of sine is cosine and the derivative
of cosine is negative sine we don't have the any derivative rules for the four remaining
sort of basic trig functions if you will so we are going to derive the derivative of secant
right now the derivative of secant okay well we don't know that yet right so we need to
write this secant in terms of its reciprocal function something we do know the derivative
with respect to x of secant of x is going to be rewritten as the derivative with respect to x
of 1 over the cosine of x and now that we have secant written into a a form that or a function
that we know how to take a derivative of the derivative of cosine is negative sine we can
start to work with this once you learn the chain rule you may work this out differently but we
don't know that yet so what i'm going to do is i'm going to recognize that the derivative with
respect to x of secant has been converted into the derivative with respect to x of the function
1 divided by the function cosine of x so following this formula this is going to be equal to
the denominator cosine of x d high the derivative with respect
with respect to x of the numerator so low d high minus high the numerator of course again 1
times d low times the derivative with respect to x of the cosine of x
all over low low all over cosine squared of x okay so what do we have here
the derivative with respect to x or the derivative really kind of regardless
of what the variable is with respect to the derivative of a constant is equal to zero right
again that slope of a horizontal line y equals one so the horizontal line is zero so we have cosine
of x times zero minus one times the cosine the derivative with respect to x of cosine of x is
going to be negative sine of x all over cosine squared x well obviously anything times zero
is zero and when you add or subtract by zero nothing happens it's the additive
identity for addition and then we have negative 1 times negative sine is positive sine
of x over cosine squared of x so the derivative of secant is equal to secant over cosine of x
but um if you already looked up the formula for uh the derivative of secant with respect to x
that's not you're not going to see this so how do we go from sine of x over cosine squared of x to
what the book says the book's sort of formula or answer for the derivative of secant does
not involve any powers of 2. so i'm going to bring this up here and we're going to write the
sine of x over cosine squared of x as 1 over cosine of x times sine of x over
cosine of x and now in this format we can see basically what you may already know from your
book if you looked at your derivatives of all six of your trig functions the derivative of secant
of x is equal to well 1 over cosine of x is secant of x and sine over cosine of course of x is equal
to tangent of x so now i'm going to erase this so i have it out of my way we have done our own work
to derive the derivative formula for the secant of x which is secant of x times tangent of x and
you can do this with the derivative of cosecant the derivative of tangent and the derivative of
cotangent all four of those trig functions their derivatives are derived through the process of the
quotient rule for derivatives isn't that cool all these formulas you're learning are not just
you know they're all related somehow and that's the great thing about math if you really uh you
don't want to try and memorize or excuse me learn math through the memorization of just a
bunch of formulas and never see how they connect because then you're just reliant on your memory
uh and even if your memory is great you never really have that deep
understanding that we're going to need to be successful in mathematics
now for our second example if the x squared plus 6 was our numerator
and our denominator was the monomial if we were taking the derivative with respect to x of x
squared plus 6x plus 6 excuse me all over 4x i would write as we've done earlier in our book when
we were studying the derivative using power rule i would give each of the terms in the numerator
that denominator and have it the derivative with respect to x of x squared over 4x plus 6 over 4x
simplify each of these terms first and then take the derivative of each term individually
using the power rule whenever your derivative denominator is a monomial and you can break it
up into separate terms it's generally going to be true for expressions that involve algebraic
expressions in the numerator and denominator not things that involve trigonometry it's
going to be easier to break that up this would i would not do the derivative of this expression
x squared plus 6 plus 6 over 4x using the quotient rule it'll work you can but i wouldn't i would
split it up and do the derivative of each term using power rule but we cannot do that
and be added in any way when our denominator is a binomial so we're going to walk through that
quotient rule the derivative of 4x over x squared plus 6 is going to be
equal to the low low the denominator x squared plus 6 d high the derivative with respect to x
of 4x high minus high how you doing d low high 4x times the derivative with respect to x of the
make denominator i keep track of what i'm saying low d high minus high d
low all over low low the denominator
squared the derivative with respect to x of 4x you're going to bring that power
of 1 which is on the x bring it down multiply it to the 4 reduce the power
on the x which is 1 reduce that to a power of 0 which of course is going to allow that x to cancel
out the derivative of 4x with respect to x is just 4. so we're looking at x squared plus 6 times 4
minus 4x times always use sub parenthesis when you do substitution the derivative with respect
to x of x squared that power of 2 comes down we reduce the power by 1 so we're looking at 2x
the derivative with respect to x of that constant 6 is going to be 0 so we just have that 2x
all over again that denominator of x squared plus 6 squared we're never going to be you know
maybe never is a strong word but we rarely want to expand out that denominator maybe a lot of
times when you're taking derivatives through the simplification process maybe some things
cancel out and you'll be able to factor out a uh cancel out a common factor of x squared plus six
maybe there's a factor that in the numerator not this one because it's not in both terms
or maybe you need to find some critical values for your derivative just don't bother
distributing out that denominator our final almost final answer here we've got distributing the 4
through the first parentheses and multiplying these together we have 4 x squared plus 24
minus 8x squared all over x squared plus six and i guess i'm writing a little bit too big here
um is going to give us a final answer after we combine like terms the derivative with respect
to x of four x over x squared plus six is going to be equal to well negative 4 x squared plus
24 all over x squared plus 6 squared and that is the end of our second
example and double check here we are good our third example is going to be
kind of similar to what you saw hopefully if you watch the video when we had the product rule
a problem that is dependent on you being able to read graphs of functions
and again just making sure we understand the difference between
a function and a derivative function so let me get the board cleaned off we'll be right back
for our third and last example as we learn the foundations the basics of these this quotient rule
we have find h let's use proper notation h prime of 8 if h of x is defined as f of x over g of x
and we don't have any functions f and g to take the derivative of
except we do have their graphs so we have here a orange graph for f of x and a green graph of g of
x and we are going to just use the picture we are not going to take the time to find the piecewise
functions that make up the whole segmented portion here of f x and g of x we are just going to
look at this problem say oh i need h prime of x i need a derivative function which is going
to tell me the slope of this function you know hopefully where i asked for and we're not going
to be asking for a slope at a sharp bend or a discontinuity or something like that so h
prime of x is equal to well the original function h is a combination of two functions f of x divided
by g of x so it's going to be low and i'm going to drop the the of x part and say low so g
d prime times the derivative of the numerator so g times f prime minus low d high minus high
f d low g prime so f times g prime all over low low or all over g squared now i'm honestly
kind of regretting i didn't include the x's in the notation so i'll do that for the next one
so we have the requestion here of finding h prime of 8. find the h prime of 8.
find the if h is a division of these two functions what is the slope of h at only at that specific
value that instantaneous rate of change when x is equal to 8. so we need to figure out
what is g of 8 multiply that by f prime of 8 minus f of 8 times g prime of 8 all over g of 8 squared
what's g of 8 what y value do you get from function g when you plug in the value of 8
well there's nothing to plug into we're going to look at the picture
we're going to go over to an x value of 8 go up or down until we hit function g
and it looks like g of 2 thinking of the answer g of 8 when i go over to the x
value of 8 plug it into g i get a y value or an answer of 2. so g of 8 is equal to 2
times f prime of 8 what is the slope of function f specifically at that x value of 8 and since
it's a linear function we have the same slope from looks like maybe 9 to infinity but if we go up 2
and over 1 we have f prime of 8 is equal to 2. now that's going to be minus f of 8 okay f prime of 8
is the slope not the same thing as f of 8 what y value are we getting from function f when we
look at the x value of 8 so go over to x equals eight go up to hit graph f of x and find that y
value of six g prime of eight what's the j what's the prime what's the slope what's the derivative
what's the slope of g when x is equal to 8. when we go over here to x equals 8 and we go find that
green basically piecewise function combination of multiple linear functions we're going to go down 1
and over 2 to hit the graph again so the slope of g when x is equal to 8 is negative one half
and i don't like to do my work horizontally but we're going to obviously have to start doing that
the again that y prime of g g of 8 is equal to 2 is going to be squared so our denominator is 2
squared and we have 2 times 2 is 4. we have minus 6 times negative 1 half is a positive 3
so we have 4 plus 3 over 2 squared is 4 so we are looking at 7 4 which is again our original answer
just kind of write this as a summary statement we have h prime of 8 is equal to 7 4 and now i hope
you have a good understanding of the use of the quotient rule i'm mr true go do your homework
you