Derivative using product rule and chain rule: y=t(ln10t)^2

[BigNate]

Hello,

I'm trying to find the derivative of y=t(ln10t)^2

I have a hint in my textbook to use the product rule and the chain rule. I see where to use the product rule, but not the chain rule. Here is what I have done:

y=t(ln10t)^2
y= 1(2ln10) + t(0) using the product rule

But this is not the right answer. Can someone please help? Thank you!

 

[Deleted member 4993]

Hello,

I'm trying to find the derivative of y=t(ln10t)^2

I have a hint in my textbook to use the product rule and the chain rule. I see where to use the product rule, but not the chain rule. Here is what I have done:

y=t(ln10t)^2
y= 1(2ln10) + t(0) using the product rule ← How did you get that

But this is not the right answer. Can someone please help? Thank you!

Chain rule is: If y = f[g(x)] then

dy/dx = [df/dg] * [dg/dx]

Is your function

y = t * [ln(10t)]²

or something else?

 

[HallsofIvy]

Hello,

I'm trying to find the derivative of y=t(ln10t)^2

I have a hint in my textbook to use the product rule and the chain rule. I see where to use the product rule, but not the chain rule. Here is what I have done:

y=t(ln10t)^2
y= 1(2ln10) + t(0) using the product rule

But this is not the right answer. Can someone please help? Thank you!

That's not the product rule. The product rule is (uv)'= u'v+ uv'.

Here, u= t and v= (ln(10t))^2.
u'= 1 and v'= 2 ln(10t)(1/t)

so u'v+ uv'= (ln(10t)^2+ 2 ln(10t).