Question Video: Differentiating Functions Involving Exponential Functions Using the Product Rule | Nagwa Question Video: Differentiating Functions Involving Exponential Functions Using the Product Rule | Nagwa

Question Video: Differentiating Functions Involving Exponential Functions Using the Product Rule Mathematics

Find the first derivative of the function 𝑦 = 2π‘₯^(8)𝑒^(5π‘₯).

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Video Transcript

Find the first derivative of the function 𝑦 equals two π‘₯ to the power of eight 𝑒 to the power of five π‘₯.

Well, we can actually look at our function and see that’s in the form 𝑦 equals 𝑒𝑣. And therefore, what we can actually use is the product rule to actually help us differentiate. And the product rule tells us that 𝑑𝑦 𝑑π‘₯ equals 𝑒 𝑑𝑣 𝑑π‘₯ plus 𝑣 𝑑𝑒 𝑑π‘₯. So therefore, what we need to do is decide in our function, if we want to find the first derivative, what 𝑒 and 𝑣 are going to be.

Well, I’ve actually decided that 𝑒 is gonna be equal to two π‘₯ to the power of eight. And 𝑣 is equal to 𝑒 to the power of five π‘₯. So now what we need to do is actually find 𝑑𝑒 𝑑π‘₯ and 𝑑𝑣 𝑑π‘₯. I’m gonna start with 𝑑𝑒 𝑑π‘₯. And I’ll do that by differentiating two π‘₯ to the power of eight, which would just give me 16π‘₯ to the power of seven. And we get that same way that we differentiate any term. We’ve multiplied the exponent by the coefficient, so eight multiplied by two, which gives us 16. And now we’d reduced the exponent by one, so from eight to seven.

Okay, now we move on to 𝑣 to find 𝑑𝑣 𝑑π‘₯. So now to find 𝑑𝑣 𝑑π‘₯, what we’re actually gonna use is a rule. And that rule states that if 𝑦 is equal to 𝑒 to the power of 𝑓 π‘₯, then 𝑑𝑦 𝑑π‘₯ is equal to the derivative of 𝑓 π‘₯ multiplied by 𝑒 to the power of 𝑓 π‘₯. And we actually get this from the chain rule, because if we have 𝑑𝑦 𝑑π‘₯ is equal to 𝑑𝑦 𝑑𝑒 multiplied by 𝑑𝑒 𝑑π‘₯, then we will have 𝑒 is equal to 𝑓 π‘₯ and 𝑦 is equal to 𝑒 to the power of 𝑒.

So therefore, we’d have that 𝑑𝑦 𝑑π‘₯ is equal to the derivative of 𝑓 π‘₯ multiplied by 𝑒 to the power of 𝑒. And that’s because the derivative of 𝑒 to the power of 𝑒 is just 𝑒 to the power of 𝑒 because 𝑒 is a number for which the gradient of 𝑒 to the power of π‘₯ is 𝑒 to the power of π‘₯. Okay, so then if we substitute back in for 𝑒, we just get the first derivative of 𝑓 π‘₯ multiplied by 𝑒 to the power of 𝑓 π‘₯.

Okay, great! So we’ve just seen where we’ve actually got this rule from. Let’s use it to actually carry on and differentiate our function. Well, we’re gonna get 𝑑𝑣 𝑑π‘₯ is equal to five multiplied by 𝑒 to the power of five π‘₯. And that’s because the derivative of five π‘₯ is just five. So we’ve got the derivative of 𝑓 π‘₯, which was five π‘₯, multiplied by 𝑒 to the power of five π‘₯.

Okay, so now at this stage, we can actually use the product rule to actually find out what 𝑑𝑦 𝑑π‘₯ or the first root of our function is going to be. So first of all, we have 𝑒, which is two π‘₯ to the power of eight, multiplied by 𝑑𝑣 𝑑π‘₯, which is five 𝑒 to the power of five π‘₯. And this is plus 𝑒 to the power of five π‘₯, because that’s our 𝑣 multiplied by our 𝑑𝑒 𝑑π‘₯, which is 16π‘₯ to the power of seven.

Okay, so we now are just gonna rearrange this to find our 𝑑𝑦 𝑑π‘₯ and just tidy up. So therefore, we’re gonna get that the first derivative of our function 𝑦 equals two π‘₯ to the power of eight 𝑒 to the power of five π‘₯ is equal to 10π‘₯ to the power of eight multiplied by five 𝑒 to the power of five π‘₯ plus 16π‘₯ to the power of seven multiplied by 𝑒 to the power of five π‘₯.

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