### Video Transcript

Find the first derivative of π¦ is equal to seven π₯ squared minus root π₯ multiplied by negative π₯ squared plus seven root π₯.

Weβre asked to find the first derivative of π¦ is equal to some function of π₯. This means we need to differentiate π¦ with respect to π₯. And we can see that weβre given π¦ as the product of two functions. So weβll do this by using the product rule for differentiation. We recall the product rule tells us if we have two functions π’ and π£, then the derivative of π’ times π£ with respect to π₯ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.

So to differentiate this by using the product rule, weβll set π’ to be equal to seven π₯ squared minus root π₯ and π£ to be equal to negative π₯ squared plus seven root π₯. So to use the product rule, weβre going to need expressions for dπ’ by dπ₯ and dπ£ by dπ₯. Letβs start with dπ’ by dπ₯. Thatβs the derivative of seven π₯ squared minus root π₯ with respect to π₯. The first thing weβll do is rewrite negative root π₯ as negative π₯ to the power of one-half by using our laws of exponents.

We can now differentiate both of these terms by using the power rule for differentiation. We multiply by the exponent of π₯ and reduce this exponent by one. This gives us two times seven times π₯ to the power of two minus one minus one-half times π₯ to the power of one-half minus one. And this simplifies to give us 14π₯ minus π₯ to the power of negative one-half divided by two. We can do the same to find dπ£ by dπ₯. Thatβs the derivative of negative π₯ squared plus seven root π₯ with respect to π₯. Again, by using our laws of exponents, weβll rewrite seven root π₯ as seven π₯ to the power of one-half.

Applying the power rule for differentiation to each term, we get negative two times π₯ to the power of two minus one plus one-half times seven π₯ to the power of one-half minus one. And we can simplify this to give us negative two π₯ plus seven π₯ to the power of negative one-half divided by two. Weβre now ready to find the first derivative of π¦ with respect to π₯ by using the product rule. Itβs equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. Substituting in our values for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯, we get the following expression for dπ¦ by dπ₯.

Weβre now ready to simplify this expression. Weβll start by multiplying out our first term. We get negative 14π₯ cubed plus 49 over two π₯ to the power of three over two plus two π₯ to the power of three over two minus seven over two. We can add 49 over two π₯ to the power of three over two and two π₯ to the power of three over two together. This gives us 53π₯ to the power of three over two divided by two. Weβll write this as 53π₯ root π₯ over two. So simplifying our first term, we got negative 14π₯ cubed plus 53π₯ root π₯ over two minus seven over two.

We now want to do the same with our second term. Doing this, we get negative 14π₯ cubed plus π₯ to the power of three over two divided by two plus 98π₯ to the power of three over two minus seven over two. And we can then add π₯ to the power of three over two over two and 98π₯ to the power of three over two to get 197π₯ to the power of three over two divided by two. And weβll write this as 197π₯ root π₯ over two. So weβve now shown dπ¦ by dπ₯ is equal to negative 14π₯ cubed plus 53π₯ root π₯ over two minus seven over two minus 14π₯ cubed plus 197π₯ root π₯ over two minus seven over two.

And we can then simplify this by combining like terms. And when we do this, we get negative 28π₯ cubed plus 125π₯ root π₯ minus seven. Therefore, for π¦ is equal to seven π₯ squared minus root π₯ times negative π₯ squared plus seven root π₯, weβve shown the first derivative of π¦ with respect to π₯ is equal to negative 28π₯ cubed plus 125π₯ root π₯ minus seven.