BAM Mr. Tarrou so you're doing your calculus homework practicing your
product and your quotient rule and then you'll end up with a problem like this
which is y is equal to X to the 3/4 power times the cube root of x squared
plus 1 all divided by 2 plus 5 all that raised to the fifth power and this is
going to be a lot a lot of work if the only tools in your toolbox
are the product rule the quotient rule and you know the power rule for finding
derivatives so what we're going to do is we're going to work through some
logarithmic differentiation we're going to let natural logs you know that
concept maybe in precalculus that you were just like logarithms they're gonna
come in really really handy right now step one we're going to take the natural
log of both sides of this equation and what that is going to allow us to do is
then on the right hand side where we have all this multiplication and
division going on that that power of 1/3 around the x squared plus 1 we're going
to be able to expand all of this into separate natural log terms so let's not
forget the natural log of a times B is equal to the natural log of a plus the
natural log of B remember logarithms give us exponents and you add exponents
when you're multiplying under a like base in this case the like base is log
base E and the natural log of a divided by B is the natural log of a minus the
natural log of B and then our third property of natural logs or really any
longer if is when you take a log of something that is raised to a power you
can bring that power out front and change it into multiplication because
remember logarithms give you exponents and you effectively have an exponent
raised to another exponent so the palm the power rule basically here is going
to say that this is equal to B times the natural log of a so we're going to
expand this second line here to be the natural
log of y is equal to the natural log of X to the 3/4 power times the cube root
of x squared plus 1 minus the natural log of 2x plus 5 raised to the fifth
power now we are in calculus I'm not going to break this up into every single
individual step actually maybe yes I will so we have now this is going to be
the natural log of y is going to be equal to the natural log of X to the 3/4
power plus the natural log of x squared plus 1 raised to the 1/3 power because
of course inside this radical we have this expression raised to the 1st so now
that expression is going to be raised to the 1/3 power right in that radical as
our fractional exponent minus the natural log of 2x plus 5 raised to the
5th power well now we're going to go ahead and just drop all these of the
power property for a run the expansion of natural logarithms and bring all
these powers out front so we have the natural log of Y is equal to 3/4 times
the natural log of X plus bringing this power down of 1/3 1/3 times the natural
log of x squared plus 1 and then minus 5 times the natural log of 2x plus 5 now
that we have taken this very complex fraction and written it as three
separate terms dealing with just natural logs we can take the derivative of each
of these individual terms remembering of course that the derivative of natural
log is or the derivative of the natural log of u is u prime over u so we're
going to have 4 here the derivative and we're taking that derivative with
respect to X so we have the derivative of the natural
log of Y is going to be equal to one over Y and n times dy/dx which you may
very well be writing as Y prime is equal to 3/4 the derivative of natural log of
X with respect to X is 1 over X now we've got 1/3 times the derivative of x
squared plus 1 is equal to 2x over our original x squared plus 1 and now we
have a little bit minus 5 the derivative of 2x plus 5 is going to be just 2 so
there's our U prime over u which is 2x plus 5 now we're not going to clean this
up too much here because the whole point is to show you how you can get to the
derivative quite quickly now whether or not this format is going to match say
multiple choice answer if this happens to be a multiple choice question is
another thing but we will just simply need to maybe just a tiny bit here we
have 1 over Y dy DX we've got 3 over 4x plus 2 x over 3x squared plus 3 or maybe
just write it as 3 parenthesis x squared plus 1 of course minus 10 over 2x plus 5
actually and that leave in that in factored form would be beneficial if you
wanted us for some reason at all three of these fractions up but now to find
out what Y prime is or dy/dx we just simply have to multiply both sides of
the equation by Y so Y prime another version of writing our derivative Y with
respect to X is equal to Y times all of this now just think of like when you're
doing implicit differentiation you're not allowed to write a second derivative
second derivative with a first derivative and the expression so I don't
really want a Y prime or dy/dx in terms of x and y's we're just going to come up
here and find that original expression for y
and substitute it in and we're going to be done y prime is equal to not a clean
answer but we got there much quicker than we would have had we just been
using the product and quotient rule that version wouldn't be clean either
very neat looking solution so we have X to the 3/4 times the cube root of x
squared plus 1 all over 2 X plus 5 squared times well all of this we have 3
over 4 X plus 2x over 3 x squared plus 3 minus 10 over 2 X plus 5 and that is
logarithmic differentiation you want to pull this out of your tool box any time
you start seeing really complex combination of multiplication and
division where you just know that man I do not want to take that derivative with
just the derivative excuse me the product or quotient rule I miss a true
BAM go to your homework
you