hi everyone today we're going to talk about how to use implicit differentiation to find
the second derivative of y to complete this problem we'll use implicit differentiation
to find the first derivative then solve for y prime and then we'll use implicit differentiation
again to find the second derivative let's take a look in this particular problem we've
been asked to use implicit differentiation to find the second derivative of y if x to
the fourth plus y to fourth is equal to sixteen remember that whenever we use implicit differentiation
we're differentiating with respect to x but we're treating y as a function and every time
we take the derivative of y we have to use chain rule and multiple by the derivative
of y assuming that it's a function so let's just go ahead and find the first derivative
of y and we'll take about what we mean as we go so we're going to be taking the derivative
of x to fourth plus y to the fourth equals sixteen with respect to x what that means
is that whenever we take the derivative of a term that has nothing but x involved in
it meaning it has no y's involved in the term then we're just taking a normal derivative
and we don't have to worry about it so x to the fourth the derivative of x to the fourth
will be four x cubed nothing special about it this is a typical derivative for us however
the derivative of y to the fourth we'll have to use chain rule and treat y as a function
so we'll treat it just like x to start out with we'll take the derivative of y to the
fourth and get four y cubed just like the derivative of x to the fourth however because
y is a function what we have to do is multiply by the derivative of y and we'll call it y
prime the easiest way to think about implicit differentiation is just to remember that whenever
you take the derivative of y you have to multiple by y prime we'll go ahead and take the derivative
of the right hand side because the derivative of any constant is just zero the derivative
of sixteen will be zero this is our first derivative we want and go ahead and solve
for y prime whenever you're asked to find the second derivative you have to find the
first derivative first and you want to go ahead and solve for y prime or the first derivative
of y before you move on to the second derivative so in order to do that we'll subtract four
x cubed from both sides and we'll get four y cubed times y prime is equal to negative
four x cubed now we'll go ahead and divide both sides by four y cubed to get y prime
on its own and we'll get y prime equals negative four x cubed divided by four y cubed we can
go ahead and cancel out our fours and we'll get negative x cubed over y cubed now to go
ahead and find the second derivative of y or y double prime we'll be taking the derivative
of negative x cubed divided by y cubed obviously because we're dealing with a quotient here
we can use the quotient rule to find the derivative so remember that the quotient rule tells us
that we're going to be taking the derivative of our numerator times the denominator minus
the derivative of the denominator times the numerator all divided by the denominator squared
so let's go through that one piece at a time we're going to be taking the derivative of
the numerator and let's just associate this negative sign here with the numerator for
now so we will take the derivative of negative x cubed and let's just go ahead and write
in each piece here we'll use d dx to signify that we want to take the derivative of the
value that comes after it so negative x cubed that is the derivative of the numerator we're
going to multiplying that by the denominator then we're going to subtract and we're going
to take the derivative of the denominator so d dx of y cubed times the numerator so
negative x cubed and then we'll divide that by the square of the denominator and our denominator
is y cubed we want to take the square of it so that's what our quotient rule formula tells
us the second derivative of y will be we just need to go ahead and take our derivatives
and simplify so the derivative of negative x cubed with respect to x again y isn't involved
here so we don't have to worry about anything fancy we just take the derivative and we get
negative three x squared we're multiplying that by y cubed then over here we have the
derivative of y cubed this will involve multiplying by the derivative of y or that chain rule
application we were talking about so when we take the derivative of y cubed what we'll
get three y squared just like if it were a normal x variable but because we're taking
the derivative of something that involves y here we have to multiply by y prime so we
get that and then we're multiplying here by negative x cubed and our denominator we have
y cubed squared which of course gives us y to the sixth so when we continue with our
simplification here we'll get negative three x squared y cubed we have a negative sign
here and a negative sign here so those will combine to become a plus sign so we'll get
plus three x cubed y squared y prime all divided by y to the sixth now we get to the reason
why we ended up solving for the first derivative of y in an earlier step we have this term
here y prime or the first derivative of y and we solved for that up here we got negative
x cubed divided by y squared we want and go ahead and plug this value in for y prime so
we'll continue simplifying we'll get y double prime equals negative three x squared y cubed
plus three x cubed y squared y squared and we'll multiply by negative x cubed over y
cubed all divided by y to the sixth now in order to simplify we need to find a common
denominator in our numerator so the first thing we'll do is we'll go ahead and simplify
that second term by multiplying the first derivative through the second term so because
we have this negative sign we'll get minus three x to the sixth because we have x cubed
and x cubed and then we have y squared divided y cubed which means that we'll get the y squared
to cancel and we'll just be left with y in the denominator of that term and then we have
this whole thing over y to the sixth so now we have y in our denominator here our denominator
of the numerator we need to find a common denominator with this term so we'll multiply
this term by y over y so that we can find a common denominator and we'll get is negative
three x squared y to the fourth because we multiplied by y over y minus three x to the
sixth which comes from this term here all divided by y so we got that common denominator
and then this whole thing is still divided by y to the sixth now we want to combine this
this denominator y here with this one y to the sixth this whole fraction in the numerator
divided by y to the sixth is the same thing as multiplying by one over y to the sixth
so what we're going to end up with is the denominator of y to seventh so we'll get negative
three x squared y to the fourth minus three x to the sixth all divided by y to the seventh
but this particular problem we can simplify further by factoring a three x squared out
of our numerator let's go ahead and factor out a actually negative three x squared and
we'll get negative three x squared times y to the fourth plus x to the fourth still all
divided by y to the seventh and that's really as far as we can simplify so that's our final
answer so I hope you found that video helpful if you did like this video down below and
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