Question Video: Differentiating Combinations of Polynomial and Root Functions Using the Product Rule at a Point

Find the first derivative of the function 𝑦 = (9π‘₯Β² βˆ’ 7)√(2π‘₯ + 1) at π‘₯ = 0.

04:51

Video Transcript

Find the first derivative of the function 𝑦 is equal to nine π‘₯ squared minus seven multiplied by the square root of two π‘₯ plus one at π‘₯ is equal to zero.

The question wants us to find the first derivative of our function 𝑦 of π‘₯ at the point where π‘₯ is equal to zero. And we can see that our function 𝑦 of π‘₯ is the product of two functions. It’s the product of nine π‘₯ squared minus seven and the square root of two π‘₯ plus one. So, to find the derivative of the product of two functions, we’ll use the product rule.

The product rule tells us if we have two functions 𝑒 and 𝑣, then the derivative of the product 𝑒 times 𝑣 is equal to 𝑣 times 𝑒 prime plus 𝑒 times 𝑣 prime. So, we’ll set our function 𝑒 of π‘₯ to be nine π‘₯ squared minus seven, and we’ll set our function 𝑣 of π‘₯ to be the square root of two π‘₯ plus one. So, to use the product rule, we need to find expressions for 𝑒 prime and 𝑣 prime. Let’s start with finding an expression for 𝑒 prime of π‘₯.

𝑒 prime of π‘₯ is the derivative of nine π‘₯ squared minus seven with respect to π‘₯. We can evaluate this by using the power rule for differentiation. This gives us that 𝑒 prime of π‘₯ is equal to 18π‘₯. We now want to find an expression for 𝑣 prime of π‘₯; however, we can’t do this directly. We see that our function 𝑣 of π‘₯ is the composition of two functions. It’s the square root of a linear function. This means to evaluate this derivative, we’re going to need to use the chain rule.

We recall the chain rule tells us if we have two functions 𝑓 and 𝑔, then the derivative of 𝑓 composed with 𝑔 of π‘₯ is equal to 𝑔 prime of π‘₯ times 𝑓 prime evaluated at 𝑔 of π‘₯. So, to use the chain rule, we’re going to need to set our function 𝑔 of π‘₯ to be our inner function. That’s the linear function two π‘₯ plus one. Then, by using this definition of 𝑔 of π‘₯, we see that 𝑣 of π‘₯ is equal to the square root of 𝑔 of π‘₯. So, we’re taking the square root of 𝑔. So, we’ll set out our function 𝑓 of 𝑔 to just be the square root of 𝑔. Using these definitions of 𝑓 and 𝑔, we’ve rewritten 𝑣 of π‘₯ to be equal to 𝑓 composed with 𝑔 of π‘₯.

We’re now ready to apply the chain rule. The chain rule tells us that 𝑣 prime of π‘₯ is equal to 𝑔 prime of π‘₯ times 𝑓 prime evaluated at 𝑔 of π‘₯. We now need to find expressions for 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. Let’s start with 𝑔 prime of π‘₯. 𝑔 of π‘₯ is a linear function. So, the derivative of 𝑔 of π‘₯ is equal to the coefficient of π‘₯ which is two. We now want to find 𝑓 prime of 𝑔. That’s the derivative of 𝑓 of 𝑔 with respect to 𝑔.

To do this, we’ll use our laws of exponents to rewrite the square root of 𝑔 as 𝑔 to the power of one-half. We can then find 𝑓 prime of 𝑔 by using the power rule for differentiation. We multiply by the exponent of 𝑔, that’s one-half, and then reduce this exponent by one. And again, by using our laws of exponents, we can rewrite one-half 𝑔 to the power of negative one-half to be equal to one divided by two root 𝑔.

Substituting these expressions into our chain rule, we’ve shown that 𝑣 prime of π‘₯ is equal to two times one divided by two times the square root of two π‘₯ plus one. And remember, we take the square root of two π‘₯ plus one because we’re evaluating 𝑓 prime of 𝑔 at 𝑔 of π‘₯. And 𝑔 of π‘₯ is two π‘₯ plus one. And we can simplify this expression slightly. We’ll cancel the shared factor two in the numerator and the denominator. So, this gives us 𝑣 prime of π‘₯ is equal to one divided by the square root of two π‘₯ plus one.

We now have all the information we need to apply the product rule to our function 𝑦 of π‘₯. This tells us that 𝑦 prime of π‘₯ is equal to 𝑣 of π‘₯ times 𝑒 prime of π‘₯ plus 𝑒 of π‘₯ times 𝑣 prime of π‘₯. Substituting in our expressions for 𝑒 of π‘₯, 𝑣 of π‘₯, 𝑒 prime of π‘₯, and 𝑣 prime of π‘₯, we have that 𝑦 prime of π‘₯ is equal to the square root of two π‘₯ plus one times 18π‘₯ plus nine π‘₯ squared minus seven multiplied by one divided by the square root of two π‘₯ plus one. We might be tempted to start simplifying at this point. However, remember, the question only wants us to find the derivative when π‘₯ is equal to zero. So, instead, we could just substitute π‘₯ is equal to zero into our expression for 𝑦 prime of π‘₯.

Substituting π‘₯ is equal to zero, we get that 𝑦 prime of zero is equal to the square root of two times zero plus one multiplied by 18 times zero plus nine times zero squared minus seven multiplied by one divided by the square root of two times zero plus one. And we can now just evaluate this expression. In our first term, we’re multiplying by zero, so our first term is equal to zero. Next, we know that nine times zero squared and two times zero are both equal to zero. So, this term simplifies to give us negative seven multiplied by one, which is equal to negative seven.

Therefore, we’ve shown the first derivative of the function 𝑦 is equal to nine π‘₯ squared minus seven times the square root of two π‘₯ plus one at π‘₯ is equal to zero is equal to negative seven.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.