### Video Transcript

Is the function π¦ equals one over
two plus π₯ a solution to the differential equation π¦ prime equals negative π¦
squared?

Remember that π¦ prime is another
way of saying dπ¦ by dπ₯, the first derivative of π¦ with respect to π₯. So, weβve been given a first order
differential equation, and we want to know whether the given function π¦ is a
solution to it. That is, we need to know whether
the function π¦ satisfies this equation.

Letβs begin then by first working
out what π¦ prime, or dπ¦ by dπ₯, is equal to for this function π¦. And to do this, we can first
express π¦ in an alternative form. We can write it as two plus π₯ to
the power of negative one. We can then find this derivative
using the general power rule, which says that if we have some function π of π₯ to
the power of π, then its derivative with respect to π₯ is equal to π multiplied by
π prime of π₯ multiplied by π of π₯ to the power of π minus one.

Here, our function π of π₯ is two
plus π₯, and our power π is negative one. So, applying the general power
rule, we have π, thatβs negative one, multiplied by the derivative of two plus π₯,
which is just one, multiplied by π of π₯. Thatβs two plus π₯, to the power of
π minus one. So, thatβs the power of negative
two. We can then rewrite this as
negative one over two plus π₯ all squared. So, we know what the left-hand side
of this differential equation would be for this function π¦.

On the right-hand side, we have
negative π¦ squared. So, thatβs the original function π¦
squared and then multiply it by negative one, which is equal to negative one over
two plus π₯ all squared. To square a fraction, we can square
the numerator and square the denominator. So, we have negative one squared,
which is one, over two plus π₯ all squared.

Now, we compare our expressions for
π¦ prime and negative π¦ squared. And we see that they are both equal
to negative one over two plus π₯ all squared. And therefore, they are indeed
equal to one another. This tells us that the function π¦
equals one over two plus π₯ does satisfy the given differential equation and,
therefore, it is a solution.