Find the derivative of the given function.y=x2ln(x2)y=x^2\ln(x^2)y=x2ln(x2)

2 x ( ln ⁡ ( x 2 ) + 1 ) 2x\left(\ln\left(x^2\right)+1\right) 2 x ( ln ( x 2 ) + 1 )

Find the derivative of the given function. y = x 2 ln ( x 2 ) y=x^2\ln(x^2) y = x 2 ln ( x 2 )

we want to find the derivative of the function y equals x squared times the natural log of x squared. Now here, since we have a function multiplying another function, we're going to have to use the product rule, and we're also going to use the chain rule. If you've not yet learned those rules, you're gonna wanna circle back to this problem after you've gone through those. For the rest of you, let's get into taking this derivative. Here, we're finding a dydx. Now remember, since we have a function multiplying another function, we're going to use the product rule. In our memory tool for the product rule, it tells us left d right plus a right d left. So I'm gonna take that left hand function, that is x squared, and multiply that by the derivative of my right hand function, the derivative of the natural log of x squared. I'm gonna have to apply the chain rule here. So this is one over that argument, which in this case is x squared, multiplied using the chain rule by the derivative of that x squared, which in this case is 2 x. Then I'm adding this together with right d left, so that's the natural log of x squared times the derivative of x squared, which is again 2 x. Now we can do some simplification here. These x squareds cancel, so all I'm left with here is 2x plus the natural log of x squared times 2x. Because I have these 2 x's in both of these terms, I can go ahead and factor it out. So this is 2x times 1 plus the natural log of x squared, and this is my final derivative here, dydx. Let us know if you have any questions here, and let's get a bit more practice.

Find the derivative of the given function. y = x 2 ln ⁡ ⁡ ( x 2 ) y=x^2\ln⁡(x^2) y = x 2 ln ⁡ ( x 2 )

2 x ( ln ⁡ ( x 2 ) + 1 ) 2x\left(\ln\left(x^2\right)+1\right) 2 x ( ln ( x 2 ) + 1 )

2 x ln ⁡ x 2 + 2 x 2x\ln x^2+\frac{2}{x} 2 x ln x 2 + x 2

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?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Find the derivative of the given function.
$y=x^2\ln(x^2)$

$2x\ln x^2+\frac{2}{x}$

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

$\frac{2}{x}$

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

Verified step by step guidance

Step 1: Recognize that the function y = x^2 ln(x^2) is a product of two functions: u = x^2 and v = ln(x^2). To find the derivative, we will use the product rule: (uv)' = u'v + uv'.

Step 2: Compute the derivative of u = x^2. Using the power rule, the derivative of x^2 is u' = 2x.

Step 3: Compute the derivative of v = ln(x^2). Using the chain rule, the derivative of ln(x^2) is v' = (1/x^2) * 2x = 2/x.

Step 4: Apply the product rule. Substitute u, u', v, and v' into the formula: y' = u'v + uv'. This becomes y' = (2x)(ln(x^2)) + (x^2)(2/x).

Step 5: Simplify the expression. Combine terms where possible: y' = 2x(ln(x^2)) + 2x. Factor out 2x to get y' = 2x(ln(x^2) + 1).

4:50m

Watch next

Master Derivatives of General Exponential Functions with a bite sized video explanation from Patrick

Start learning

Derivatives of Exponential & Logarithmic Functions practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice