Video: Differentiating Root and Exponential Functions

Differentiate the function 𝑦 = 3𝑒^(π‘₯) βˆ’ 5/βˆ›π‘₯.

02:42

Video Transcript

Differentiate the function 𝑦 equals three 𝑒 to the power of π‘₯ minus five over the cube root of π‘₯.

The first thing we’re gonna do is actually rewrite our equation using exponent rules. The first rule we’re gonna use is that π‘₯ to the power of one over π‘Ž is equal to the π‘Žth root of π‘₯, which means that we can actually rewrite our function as 𝑦 is equal to three 𝑒 to the power of π‘₯ minus five over π‘₯ to the power of a third.

And then we can actually use a second exponent rule, which tells us that π‘₯ to the power of negative π‘Ž is equal to one over π‘₯ to the power of π‘Ž. So, therefore, again, we can rewrite our function. And now we can rewrite it as 𝑦 is equal to three 𝑒 to the power of π‘₯ minus five π‘₯ to power of negative a third. And we’ve got that point by applying each of our exponent rules that we talked about.

You could actually go straight from the original function to the final line that we have at the moment just by applying them both in one go. I just showed you how to do it step by step so you understand how they’re used. Okay, great! So now let’s get on and differentiate our function.

So if we’re gonna find 𝑑𝑦 𝑑π‘₯, or we’re gonna find the differential of our function, the first term we’re gonna look at is this one. Our first term is three 𝑒 to the power of π‘₯. And in this term, we can actually see that we have 𝑒 to the power of π‘₯. And 𝑒 is just a number where the slope of 𝑒 to the power of π‘₯ is equal to 𝑒 to the power of π‘₯. So therefore, if we differentiate our term, we’re gonna be left with three 𝑒 to the power of π‘₯ because it doesn’t change. And this relationship can actually be proved using limits.

So now I’m gonna move on to the second term. And we’re gonna get plus five over three. And we get that because if we get negative five, which is our coefficient, multiplied by our exponent, which is negative a third, we’re gonna get positive five over three. And then this is gonna be multiplied by π‘₯ to the power of negative four over three. And we get that because if we have negative one over three and then we subtract one or three over three, we get negative four over three.

So therefore, we can say that if we differentiate the function 𝑦 equals three 𝑒 to the power of π‘₯ minus five over the cube root of π‘₯, we’re gonna get three 𝑒 to the power of π‘₯ plus five over three π‘₯ to the power of negative four over three.

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