Video Transcript
Find the first derivative of 𝑦 is equal to seven 𝑥 squared minus root 𝑥 multiplied by negative 𝑥 squared plus seven root 𝑥.
We’re asked to find the first derivative of 𝑦 is equal to some function of 𝑥. This means we need to differentiate 𝑦 with respect to 𝑥. And we can see that we’re given 𝑦 as the product of two functions. So we’ll do this by using the product rule for differentiation. We recall the product rule tells us if we have two functions 𝑢 and 𝑣, then the derivative of 𝑢 times 𝑣 with respect to 𝑥 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥.
So to differentiate this by using the product rule, we’ll set 𝑢 to be equal to seven 𝑥 squared minus root 𝑥 and 𝑣 to be equal to negative 𝑥 squared plus seven root 𝑥. So to use the product rule, we’re going to need expressions for d𝑢 by d𝑥 and d𝑣 by d𝑥. Let’s start with d𝑢 by d𝑥. That’s the derivative of seven 𝑥 squared minus root 𝑥 with respect to 𝑥. The first thing we’ll do is rewrite negative root 𝑥 as negative 𝑥 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 𝑥 and reduce this exponent by one. This gives us two times seven times 𝑥 to the power of two minus one minus one-half times 𝑥 to the power of one-half minus one. And this simplifies to give us 14𝑥 minus 𝑥 to the power of negative one-half divided by two. We can do the same to find d𝑣 by d𝑥. That’s the derivative of negative 𝑥 squared plus seven root 𝑥 with respect to 𝑥. Again, by using our laws of exponents, we’ll rewrite seven root 𝑥 as seven 𝑥 to the power of one-half.
Applying the power rule for differentiation to each term, we get negative two times 𝑥 to the power of two minus one plus one-half times seven 𝑥 to the power of one-half minus one. And we can simplify this to give us negative two 𝑥 plus seven 𝑥 to the power of negative one-half divided by two. We’re now ready to find the first derivative of 𝑦 with respect to 𝑥 by using the product rule. It’s equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. Substituting in our values for 𝑢, 𝑣, d𝑢 by d𝑥, and d𝑣 by d𝑥, we get the following expression for d𝑦 by d𝑥.
We’re now ready to simplify this expression. We’ll start by multiplying out our first term. We get negative 14𝑥 cubed plus 49 over two 𝑥 to the power of three over two plus two 𝑥 to the power of three over two minus seven over two. We can add 49 over two 𝑥 to the power of three over two and two 𝑥 to the power of three over two together. This gives us 53𝑥 to the power of three over two divided by two. We’ll write this as 53𝑥 root 𝑥 over two. So simplifying our first term, we got negative 14𝑥 cubed plus 53𝑥 root 𝑥 over two minus seven over two.
We now want to do the same with our second term. Doing this, we get negative 14𝑥 cubed plus 𝑥 to the power of three over two divided by two plus 98𝑥 to the power of three over two minus seven over two. And we can then add 𝑥 to the power of three over two over two and 98𝑥 to the power of three over two to get 197𝑥 to the power of three over two divided by two. And we’ll write this as 197𝑥 root 𝑥 over two. So we’ve now shown d𝑦 by d𝑥 is equal to negative 14𝑥 cubed plus 53𝑥 root 𝑥 over two minus seven over two minus 14𝑥 cubed plus 197𝑥 root 𝑥 over two minus seven over two.
And we can then simplify this by combining like terms. And when we do this, we get negative 28𝑥 cubed plus 125𝑥 root 𝑥 minus seven. Therefore, for 𝑦 is equal to seven 𝑥 squared minus root 𝑥 times negative 𝑥 squared plus seven root 𝑥, we’ve shown the first derivative of 𝑦 with respect to 𝑥 is equal to negative 28𝑥 cubed plus 125𝑥 root 𝑥 minus seven.
