Find the derivative of the given function.g(x)=ex2lnxg(x)=e^{x^2\ln x}g(x)=ex2lnx

x e x 2 ln ⁡ x ( 2 ln ⁡ x + 1 ) xe^{x^2\ln x}\left(2\ln x+1\right) x e x 2 l n x ( 2 ln x + 1 )

Find the derivative of the given function. g ( x ) = e x 2 ln x g(x)=e^{x^2\ln x} g ( x ) = e x 2 l n x

Here, we wanna find the derivative of the function g of x is equal to e to the power of x squared times the natural log of x. Now here, we're going to have to utilize the chain rule. Most of you should already know the chain rule, so we can continue as normal. But if you have not learned the chain rule yet, you're gonna wanna circle back to this problem once you've learned it. For the rest of you, let's get into finding our derivative, g prime of x. So starting with our outermost function, this is e to the power of this x squared times the natural log of x. We know that the derivative of our basic exponential function e to the x is just going to be e to the x. But here, since that power is not just x, it is x squared times the natural log of x, I need to put all of that still in my power. Now multiplying by the derivative of this x squared natural log of x, that's what our chain rule tells us to do. Because here, I have a function multiplying another function, x squared times the natural log of x, I need to apply the product rule. So we're not only using the chain rule. We're also using the product rule within this chain rule. So multiplying here by the derivative of x squared natural log of x. Remember our product rule memory tool here, left d right plus right d left, so that left hand function x squared times the derivative of that right hand function, which is one over x, then plus right d left, that is the natural log of x, times the derivative of x squared, that is just 2 x. Now the all of our calculus is done here. We've taken our derivative. We need to simplify this a little bit because there are some things that will cancel. Now we have this x and this x squared, and then we have 2x times the natural log of x. Let's write this a little bit neater. That is e to the x squared times the natural log of x times x plus 2 X natural log of X. Now both of these terms have an X in them, so we can go ahead and factor that out as well just to fully simplify this. This is X times e to the X squared times the natural log of X times 1 plus 2 natural log of x. And this is our final derivative here, having applied all of the rules that we know for derivatives. Let us know if you have any questions, and let's keep going.

Find the derivative of the given function. g ( x ) = e x 2 ln ⁡ x g(x)=e^{x^2\ln x} g ( x ) = e x 2 l n x

x e x 2 ln ⁡ x ( 2 ln ⁡ x + 1 ) xe^{x^2\ln x}\left(2\ln x+1\right) x e x 2 l n x ( 2 ln x + 1 )

e x 2 ln ⁡ x e^{x^2\ln x} e x 2 l n x

e 2 x + 1 x e^{2x+\frac{1}{x}} e 2 x + x 1

2 e x 2 ln ⁡ x 2e^{x^2\ln x} 2 e x 2 l n x

4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Business Calculus

4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Struggling with Business Calculus?

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Multiple Choice

Find the derivative of the given function.
$g(x)=e^{x^2\ln x}$

$e^{x^2\ln x}$

$e^{2x+\frac{1}{x}}$

$xe^{x^2\ln x}\left(2\ln x+1\right)$

$2e^{x^2\ln x}$

Verified step by step guidance

Step 1: Recognize that the function g(x) = e^{x^2 \ln x} is a composite function. To differentiate it, we will use the chain rule. The chain rule states that if y = f(u) and u = g(x), then dy/dx = f'(u) * g'(x).

Step 2: Start by differentiating the outer function e^u, where u = x^2 \ln x. The derivative of e^u with respect to u is e^u. So, the derivative of g(x) with respect to x will include e^{x^2 \ln x}.

Step 3: Now, differentiate the inner function u = x^2 \ln x. This requires the product rule because u is a product of x^2 and \ln x. The product rule states that if u = v * w, then u' = v'w + vw'.

Step 4: Apply the product rule to u = x^2 \ln x. Here, v = x^2 and w = \ln x. The derivative of v = x^2 is v' = 2x, and the derivative of w = \ln x is w' = 1/x. Substituting into the product rule, we get u' = (2x)(\ln x) + (x^2)(1/x). Simplify this to u' = 2x \ln x + x.

Step 5: Combine the results from Step 2 and Step 4. The derivative of g(x) is g'(x) = e^{x^2 \ln x} * (2x \ln x + x). Factor out x from the second term to write the final expression as g'(x) = x e^{x^2 \ln x} (2 \ln x + 1).

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