How do you find the first and second derivative of #x(lnx)^2#?

Question

Paisley Johnson · Accepted Answer

#f'(x) = lnx(lnx+2)#
#f''(x) = 2/x(lnx+1)# #f(x) = xln^2x# #f'(x) = x*2lnx*1/x + ln^2x*1# [Product rule, chain rule and power rule] #= 2lnx+ln^2x# #= lnx(lnx+2)# #f''(x) = lnx*d/dx(lnx+2) +d/dx lnx* (lnx+2)# [Product rule] #= lnx(1/x+0) + 1/x*(lnx+2)# [Standard differential] #= 1/x*(2lnx+2)# #= 2/x(lnx+1)#

Ezra Adams · Answer

Use the product rule to get: #y'=2ln(x)+ln(x)^2#
and the same for getting #y''=2/x(ln(x)+1)# Step 1. Break the original into two functions #u=x# and #v=ln(x)^2# Step 2. Differentiate each #u# and #v# #d/dx(u)=d/dx(x)=1# #d/dx(v)=d/dx(ln(x)^2)=2*ln(x)*1/x# Step 3. Use the Product Rule to solve Product Rule: #dy/dx=u (dv)/dx+v (du)/dx# Plugging in: #dy/dx=x*2ln(x)1/x+ln(x)^2*1# #dy/dx=2ln(x)+ln(x)^2=ln(x)(2+ln(x))# Step 4. Differentiate the answer from Step 3. #(d^2y)/dx^2=ln(x)1/x+(2+ln(x))1/x# #=ln(x)/x+2/x+ln(x)/x# #=1/x(2ln(x)+2)# #=2/x(ln(x)+1)#

Scarlett Allison · Answer

undefined To find the first derivative of $ x(\ln x)^2 $, apply the product rule. The result is $ (2\ln x + 1)(\ln x)^2 + 2x\ln x/x $.

To find the second derivative, differentiate the first derivative obtained above using the product rule and simplify. The result is $ \frac{2}{x}(1 + 2\ln x) + 2(\ln x)^2 + \frac{2}{x} $.

Isabella Ackerman · Answer

undefined To find the first and second derivatives of $ x(\ln(x))^2 $, we will use the product rule and the chain rule.

Let $ f(x) = x $ and $ g(x) = (\ln(x))^2 $.

The first derivative of $ f(x) $ is $ f'(x) = 1 $.

The first derivative of $ g(x) $ can be found using the chain rule:

$$
g'(x) = 2(\ln(x)) \cdot \frac{1}{x} = \frac{2\ln(x)}{x}
$$

Now, applying the product rule:

$$
\frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)
$$

$$
\frac{d}{dx}[x(\ln(x))^2] = 1 \cdot (\ln(x))^2 + x \cdot \frac{2\ln(x)}{x}
$$

$$
\frac{d}{dx}[x(\ln(x))^2] = (\ln(x))^2 + 2\ln(x)
$$

This is the first derivative. Now, let's find the second derivative.

The derivative of $ (\ln(x))^2 $ can be found using the chain rule again:

$$
\frac{d}{dx}[(\ln(x))^2] = 2\ln(x) \cdot \frac{1}{x} = \frac{2\ln(x)}{x}
$$

Now, the second derivative using the product rule:

$$
\frac{d^2}{dx^2}[x(\ln(x))^2] = \frac{d}{dx}[(\ln(x))^2 + 2\ln(x)]
$$

$$
\frac{d^2}{dx^2}[x(\ln(x))^2] = 2\cdot \frac{1}{x} + 2\cdot \frac{1}{x} = \frac{2}{x} + \frac{2}{x} = \frac{4}{x}
$$

So, the first derivative of $ x(\ln(x))^2 $ is $ (\ln(x))^2 + 2\ln(x) $, and the second derivative is $ \frac{4}{x} $.