Video Transcript
Find the first derivative of the function 𝑦 equals 𝑥 to the fifth power times 𝑥 squared plus two times three 𝑥 cubed plus three 𝑥 plus six.
Our function is itself the product of three differential function. Remember, each function itself is a polynomial. And we know that these are differentiable. So we could look to apply the product rule twice. Alternatively, though, it’s just as easy to distribute 𝑥 to the fifth power over 𝑥 squared plus two so that we have the product of just two differentiable functions. So that’s 𝑥 to the fifth power times 𝑥 squared, which is 𝑥 to the seventh power. And then, we multiply 𝑥 to the fifth power by two. And we get two 𝑥 to the fifth power.
And so we can rewrite our function as 𝑦 equals 𝑥 to the seventh power plus two 𝑥 to the fifth power all multiplied by three 𝑥 cubed plus three 𝑥 plus six. And the reason we were allowed to do this is because multiplication is commutative. It can be done in any order. We’re next going to recall the definition for the product rule. This says that given two differentiable functions, 𝑢 and 𝑣, the derivative of their product, 𝑢 times 𝑣, is 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥.
We’re going to let 𝑢 be equal to 𝑥 to the seventh power plus two 𝑥 to the fifth power. And we’ll let 𝑣 be equal to three 𝑥 cubed plus three 𝑥 plus six. We see that to apply the product rule, we’re going to need to calculate d𝑣 by d𝑥 and d𝑢 by d𝑥. And so we recall that to differentiate a polynomial term of the form 𝑎𝑥 to the 𝑛th power, we multiply the entire term by 𝑛 and then reduce 𝑛 by one. And so, the first derivative of 𝑥 to the seventh power is seven times 𝑥 to the sixth power. And when we differentiate two 𝑥 to the fifth power, we get five times two 𝑥 to the fourth power, which can be simplified to 10𝑥 to the fourth power.
The first derivative of three 𝑥 cubed is three times three 𝑥 squared, which is nine 𝑥 squared. And then, the first derivative of three 𝑥 is simply three. Remember, when we differentiate a constant, we get zero. So now that we found d𝑢 by d𝑥 and d𝑣 by d𝑥, let’s substitute everything we have into our formula for the product rule. It’s 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥.
Our next job is to distribute our parentheses. That’s quite straightforward in the case of this first expression. It’s 𝑥 to the seventh power times nine 𝑥 squared, which is nine 𝑥 to the ninth power. Remember, we simply add the exponents. And then, when we multiply 𝑥 to the seventh power by three, we get three 𝑥 to the seventh power. Two 𝑥 to the fifth power times nine 𝑥 squared is 18𝑥 to the seventh power. And then, we multiply the last two terms. And we get six 𝑥 to the fifth power.
We’ll need to be a little bit careful with multiplying our final two expressions. Let’s begin by multiplying three 𝑥 cubed by seven 𝑥 to the sixth power to get 21𝑥 to the ninth power. Then, we’ll multiply three 𝑥 by seven 𝑥 to the sixth power and six by seven 𝑥 to the sixth power. Next, we multiply three 𝑥 cubed by 10𝑥 to the fourth power. And we get 30𝑥 to the seventh power. We multiply three 𝑥 by 10𝑥 to the fourth power. And finally, we multiply six by 10𝑥 to the fourth power. And that gives us 60𝑥 to the fourth power. Let’s collect like terms. We see we have 30𝑥 to the ninth power plus 72𝑥 to the seventh power, 42𝑥 to the sixth power, 36𝑥 to the fifth power, and finally 60𝑥 to the fourth power.
And we have the first derivative of our function. d𝑦 by d𝑥 is equal to 30𝑥 to the ninth power plus 72𝑥 to the seventh power plus 42𝑥 to the sixth power plus 36𝑥 to the fifth power plus 60𝑥 to the fourth power.
