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

Find the derivative of the function 𝑦 = 5π‘₯Β² 𝑒^(1/π‘₯).

04:12

Video Transcript

Find the derivative of the function 𝑦 equals five π‘₯ squared 𝑒 to the power of one over π‘₯.

So the first thing we can do, if we’re gonna try and differentiate our function, is take out the constant term because this won’t affects the differentiation. So, we’ve got the five. So, we can rewrite our function as the derivative of 𝑦 is equal to five, because as we said, we’ve taken the constant term outside, multiplied by the derivative of π‘₯ squared multiplied by 𝑒 to the power of one over π‘₯.

So now, what we need to do in order that we can differentiate π‘₯ squared multiplied by 𝑒 to the power of one over π‘₯ is use the product rule. Now, the product rule tells us is that if we have 𝑓 of π‘₯ multiplied by 𝑔 of π‘₯, that we can call 𝑓 of π‘₯ 𝑒, 𝑔 of π‘₯ 𝑣, and then we can say the derivative of 𝑒𝑣 is gonna be equal to 𝑒d𝑣 dπ‘₯ plus 𝑣d𝑒 dπ‘₯. So, it’s one of our terms multiplied by the derivative of the other term plus the other term multiplied by the derivative of our first term. Okay so, now we have this product rule. Let’s apply it.

So, what we’re gonna do is call our 𝑒 π‘₯ squared and our 𝑣 𝑒 to the power of one over π‘₯. And therefore, if we differentiate π‘₯ squared, so we find d𝑒 dπ‘₯, this is gonna be two π‘₯. And that’s because what we do is we multiply the exponent by the coefficient, so in this case two multiplied by one. And then, what we do is we reduce the exponent by one. So, it goes from two to one, so we’re left with two π‘₯.

And now, what we need to do is differentiate 𝑒 to the power of one over π‘₯. And in order to do that, what we’re gonna have to do is use an exponential function rule for a differentiation. And that tells us if we want to find the derivative of 𝑒 to the power of 𝑒 of π‘₯, then this is equal to 𝑒 to the power of 𝑒 of π‘₯ multiplied by the derivative of 𝑒 of π‘₯. So therefore, in our case, it’s gonna be d𝑣 dπ‘₯ is equal to 𝑒 to the power of one over π‘₯ multiplied by the derivative of one over π‘₯.

So, to differentiate one over π‘₯, what we’re gonna do is use the reciprocal rule. And, this tells us if we want the derivative of one over 𝑒 of π‘₯, this is equal to negative and then the derivative of 𝑒 of π‘₯ over 𝑒 of π‘₯ squared. So, what this is gonna give us is 𝑒 to the power of one over π‘₯, that’s what we already had. And then for the derivative of one over π‘₯, we’re gonna have negative then we’ve got one as the numerator. And that’s because the derivative of π‘₯ is just one and then over π‘₯ squared, because that’s our 𝑒 of π‘₯ squared.

So therefore, we’re gonna get negative 𝑒 to the power of one over π‘₯ over π‘₯ squared. So then, what we do is we apply our product rule. So, we’re gonna get the derivative of π‘₯ squared multiplied by 𝑒 to the power of one over π‘₯ is equal to. Then, we’ve got our 𝑒d𝑣 dπ‘₯. So, π‘₯ squared multiplied by negative 𝑒 to the power of one over π‘₯ over π‘₯ squared plus then we’ve got our 𝑣d𝑒 dπ‘₯. So, it’s 𝑒 to the power of one over π‘₯ multiplied by two π‘₯.

So, we can simplify. So, we’re gonna see that our π‘₯ squareds are going to cancel on the left-hand term. That’s cause we got π‘₯ squared multiplied by something over π‘₯ squared. So therefore, we’re gonna be left with negative 𝑒 to the power of one over π‘₯ plus two π‘₯. Multiplied by 𝑒 to the power of one over π‘₯.

So now, if we put that back into our original expression, we’ve got the derivative of 𝑦 is equal to five. Multiplied by then we’ve got two π‘₯𝑒 to the power of one over π‘₯ minus 𝑒 to the power of one over π‘₯.

But have we finished? Well, no. Because we can simplify a step further. Because what we can do is see that each of our terms has a common factor. And that factor is 𝑒 to the power of one over π‘₯. So therefore, we can take this outside our parentheses. So therefore, what we can say is that the derivative of the function 𝑦 equals five π‘₯ squared multiplied by 𝑒 to the power of one over π‘₯ is five 𝑒 to the power of one over π‘₯ multiplied by two π‘₯ minus one.

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