Question Video: Differentiating a Combination of Root and Polynomial Functions Using the Product Rule

Find the first derivative of 𝑦 = (7π‘₯Β² βˆ’ √(π‘₯))(βˆ’π‘₯Β² + 7√(π‘₯)).

03:55

Video Transcript

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

We’re asked to find the first derivative of 𝑦 is equal to some function of π‘₯. This means we need to differentiate 𝑦 with respect to π‘₯. 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 𝑒 and 𝑣, then the derivative of 𝑒 times 𝑣 with respect to π‘₯ is equal to 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯.

So to differentiate this by using the product rule, we’ll set 𝑒 to be equal to seven π‘₯ squared minus root π‘₯ and 𝑣 to be equal to negative π‘₯ squared plus seven root π‘₯. So to use the product rule, we’re going to need expressions for d𝑒 by dπ‘₯ and d𝑣 by dπ‘₯. Let’s start with d𝑒 by dπ‘₯. That’s the derivative of seven π‘₯ squared minus root π‘₯ with respect to π‘₯. The first thing we’ll 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 d𝑣 by dπ‘₯. That’s the derivative of negative π‘₯ squared plus seven root π‘₯ with respect to π‘₯. Again, by using our laws of exponents, we’ll 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. We’re now ready to find the first derivative of 𝑦 with respect to π‘₯ by using the product rule. It’s equal to 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. Substituting in our values for 𝑒, 𝑣, d𝑒 by dπ‘₯, and d𝑣 by dπ‘₯, we get the following expression for d𝑦 by dπ‘₯.

We’re now ready to simplify this expression. We’ll 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. We’ll 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 we’ll write this as 197π‘₯ root π‘₯ over two. So we’ve now shown d𝑦 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 𝑦 is equal to seven π‘₯ squared minus root π‘₯ times negative π‘₯ squared plus seven root π‘₯, we’ve shown the first derivative of 𝑦 with respect to π‘₯ is equal to negative 28π‘₯ cubed plus 125π‘₯ root π‘₯ minus seven.

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