# A little confused

#### vanillasnow

book states given ln(3x+1) h(x)e^-3x G(x)
3^x

find d/DC fx+he+gx

then find d/Dx x^2 (f(x)

how how do I solve this?

doni get the derivative to all and then add to the first problem?

for second: do I get the derivative of f(X) and multiply to x^2 OR do I take derivative to x^2 before multiplying?

#### Debsta

MHF Helper
I can't make sense of your question to help you. Could you make it clearer. Use brackets if necessary.

#### skeeter

MHF Helper
book states given ln(3x+1) h(x)e^-3x G(x)
3^x

find d/DC fx+he+gx

then find d/Dx x^2 (f(x)
a little confused? ... so am I

Did you forget some equal signs?

Is $f(x)=\ln(3x+1)$ ?

Is $h(x)=e^{-3x}$ ?

Is $g(x)=3^x$

What does d/DC mean? Do you really mean $\dfrac{d}{dx} \bigg[f(x)+h(x)+g(x)\bigg]$ ?

If so, then the derivative of a sum of functions is the sum of their derivatives ... $f'(x)+h'(x)+g'(x)$

Is the second problem $\dfrac{d}{dx} \bigg[x^2 \cdot f(x) \bigg]$ ?

If so, use the product rule ... I'm sure you were taught that since you've been asked to do it.

In future, please proofread your post by using the "preview" button, make any corrections, preview again, then submit. You'll find folks here are more inclined to help if they can read and understand what you posted. Avoid using phone text abbreviations. Thank you for your cooperation.

#### vanillasnow

First of all,

Thank you for the help. Secondly, my computer would not work and as you mentioned, was done one the phone. Unfortunately I was not able to make corrections yesterday and with the limited view screen, It was pretty hard to proof read as everything I typed had a time delay.

But yes, you retyped correct, and it was what I was asking.

Now to your question: product rule for second problem.
You'd think we ( the entire) class would know, but unfortunately we are faced with a math teacher who gives very little explanation.

In other words, when one looks at the problem at hand,problem 2- in where you pointed to use the product rule, do we first take the derivative of f'(x) and plug it (multiply it by (x^2) OR take the derivative of (x^2) as well and use the product rule...?

Thank you again, Jen

#### skeeter

MHF Helper
Product rule ...

$\dfrac{d}{dx} \bigg[u(x) \cdot v(x)\bigg] = u(x) \cdot v'(x) + u'(x) \cdot v(x)$

$\dfrac{d}{dx} \bigg[x^2 \cdot f(x)\bigg] = x^2 \cdot f'(x) + 2x \cdot f(x)$

#### vanillasnow

Thank you !!! you are excellent