Derivatives of Exponential Functions \u0026 Logarithmic Differentiation Calculus lnx, e^2x, x^x, x^sinx
Derivatives of Exponential Functions \u0026 Logarithmic Differentiation Calculus lnx, e^2x, x^x, x^sinx

Video Transcript

Determine the derivative of 𝑦
equals 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 squared.

Now to enable us to actually
determine the derivative, what we’re gonna use is the product rule. And we can actually use the product
rule when we have our function in the form 𝑦 equals 𝑢𝑣. And if we take a look at our
function, it’s actually in this form. So how does the actual product rule
work? Well the product rule tells us that
the derivative is equal to 𝑢 𝑑𝑣 𝑑𝑥 plus 𝑣 𝑑𝑢 𝑑𝑥.

So what this means is 𝑢 multiplied
by the derivative of 𝑣 plus 𝑣 multiplied by the derivative of 𝑢. Okay, great! So now that we know the product
rule, let’s use it to actually determine our derivative. So in order to actually determine
our derivative, first of all what we need to do is to decide what 𝑢 and 𝑣 are. So 𝑢 is gonna be equal to 𝑒 to
the power of negative five 𝑥, and 𝑣 is gonna be equal to 𝑥 squared.

Next, what we want to do is
actually differentiate our 𝑢 and 𝑣. So I’m gonna start with 𝑣 because
this is more straightforward. So we can say that d𝑣 d𝑥 is going
to be equal to the derivative of 𝑥 squared. Well this is gonna be equal to two
𝑥, just remind us how we did that. So our exponent multiplied by our
coefficient, so two multiplied by one, and then it’s 𝑥 to the power of, and then we
reduce our exponent by one, so two minus one, which just be one. So we get two 𝑥.

So now we can move on to 𝑢. So if we wanna find d𝑢 d𝑥, well
we’re actually gonna have to use is a general rule to help us here as well. And that’s because 𝑢 is in the
form 𝑦 is equal to 𝑒 to the power of 𝑓 of 𝑥. And our rule tells us, if we have
it in this form, then what we have is that the derivative is going to be equal to
the derivative of 𝑓 of 𝑥 multiplied by 𝑒 to the power of 𝑓 of 𝑥. And this actually comes from an
adaptation of the chain rule.

Okay, great! So we can use this to actually find
out what d𝑢 d𝑥 is going to be. Well first of all, it is gonna be
negative five. And that’s because if you
differentiate negative five 𝑥, you get negative five. And then this is gonna be
multiplied by 𝑒 to the power of negative five 𝑥. So great we now have d𝑢 d𝑥 and
d𝑣 d𝑥. So now what we can do is actually
go back to our product rule to actually find the derivative of our function.

So first of all, we’re gonna have
𝑒 to the power of negative five 𝑥 because that’s our 𝑢. And then this is gonna be
multiplied by two 𝑥. And that’s because this is our 𝑑𝑣
𝑑𝑥. And then this is gonna be plus 𝑥
squared, which is our 𝑣, and then multiplied by negative five 𝑒 to the power of
negative five 𝑥 because this is our 𝑑𝑢 𝑑𝑥. Okay, great! So now let’s rearrange this. And when we do that, we get two 𝑒
to the power of negative five 𝑥 multiplied by 𝑥 minus five 𝑒 to the power of
negative five 𝑥 multiplied by 𝑥 squared.

Okay, so now what we can do is
actually simplify this by taking out factors. So when we do that, there’s
actually gonna be two results that we can have. So I’m gonna give you both of
those. So the first one that we can find
is if we take out 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 as a factor
of each of our terms. So if we do that, then inside the
parentheses we’re gonna get two minus five 𝑥. And this is actually our derivative
simplified fully. So therefore, we can actually say
that the derivative of 𝑦 equals 𝑒 to the power of negative five 𝑥 multiplied by
𝑥 squared is 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 multiplied by two
minus five 𝑥.

Okay, as I said, there’s actually
another way that we can actually leave our answer. And we get this if we take out
negative 𝑒 to the power of negative five 𝑥 multiplied by 𝑥 as a factor this time
instead. And we do that so that we can
actually have the 𝑥 term as the first term in our parentheses this time. So what we get is the derivative is
actually equal to negative 𝑒 to the power of negative five 𝑥 multiplied by 𝑥,
then multiplied by five 𝑥 minus two. Okay, great! So we’ve actually determined the
derivative of our function. And we did that using the product
rule and then an adaptation of the chain rule.

You are watching: Question Video: Differentiating a Combination of Functions Involving Exponential Functions Using the Product Rule. Info created by Bút Chì Xanh selection and synthesis along with other related topics.