problem on novanet

[marshall21]

Hi guys, I'm new here. I am taking Novanet calculus on the computer at school because my school does not offer calculus. I have asked my math teacher for help on this problem but he says he hasn't had CALC. for 10 years and he doesn't know. I have tried setting the problem up the way he said to and I still get it wrong. Any help at all is appreciated. Each day the problem changes for me if I do not get it right, so if u could answer it before 3:30 central time, i'd appreciate it.

Well here it is:

(x+3)^4 (2x-1)^9
over
(3x+1)^3

The question asks me to use differentiation rules in combination
thanks

 

[stapel]

(x+3)^4 (2x-1)^9
over
(3x+1)^3

The question asks me to use differentiation rules in combination

Just checking: Is the function the following...?


. . . . .\(\displaystyle \large{y\,=\,\frac{(x\,+\,3)^4\,(2x\,-\,1)^}{(3x\,+\,1)^3}}\)


And are the instructions to "differentiate the function"...?

If so, I would start by using the Quotient Rule (which, when differentiating the numerator, will require the use of the Product Rule). Have you learned either of those rules?

. . .\(\displaystyle \large{\mbox{Quotient Rule:}}\)

. . . . .\(\displaystyle \large{\mbox{For }f(x)\,= \,\frac{g(x)}{h(x)}\mbox{, we have}}\)

. . . . .\(\displaystyle \large{f'(x)\,=\,\frac{g'(x)h(x)\,-\,g(x)h'(x)}{[h(x)]^2}\)


. . .\(\displaystyle \large{\mbox{Product Rule:}}\)

. . . . .\(\displaystyle \large{\mbox{For }f(x)\,= \,g(x)h(x)\mbox{, we have}}\)

. . . . .\(\displaystyle \large{f'(x)\,=\,g'(x)h(x)\,+\,g(x)h'(x)}\)


When you reply, please include all of the steps you have tried thus far.

Thank you.

Eliz.

 

[marshall21]

...I would start by using the Quotient Rule (which, when differentiating the numerator, will require the use of the Product Rule). Have you learned either of those rules?

Yes, I have learned those rules, and I have used both, but I am unable to come up wit the correct answer. Also, yes, I am supposed to differentiate the function.

 

[stapel]

Yes i have learned those rules, and i have used both but i am unable to come up wit the correct answer.

Please reply showing your steps, and providing the correct, or at least expected, answer.

Thank you.

Eliz.

 

[marshall21]

ok i gotcha.

I took the (derivative of the top) X (denominator) - (numerator X deriv. of denom)

over

denom squared

i am this far

(3x+1) [ (x+3)^7 (2x+1)^8 (18) + (2x-1)^9 (x+3)^6 (7)] - (x+3)^7 (2x-1)^9 (9)

over

(3x+1)^4


i just need to simplify now.

 

[pka]

Have you had logarithmic differentiation?
\(\displaystyle \L
\begin{array}{l}
y = \frac{{\left( {x + 3} \right)^4 \left( {2x - 1} \right)}}{{\left( {3x + 1} \right)^3 }} \\
\ln (y) = 4\ln (x + 3) + \ln (2x - 1) - 3\ln (3x + 1) \\
\frac{{y'}}{y} = ? \\
\end{array}\)

Using implicit differentiation you can finish it.

 

[marshall21]

Have you had logarithmic differentiation?
\(\displaystyle \L
\begin{array}{l}
y = \frac{{\left( {x + 3} \right)^4 \left( {2x - 1} \right)}}{{\left( {3x + 1} \right)^3 }} \\
\ln (y) = 4\ln (x + 3) + \ln (2x - 1) - 3\ln (3x + 1) \\
\frac{{y'}}{y} = ? \\
\end{array}\)

Using implicit differentiation you can finish it.

i've haven't learned that yet.

 

[marshall21]

(x+3)^6 (2x-1)^8 ( -18x^2 - 13x-20)

over

(3x+1)^4

that was my answer and it wasn't right, i have a few more chances to get it right

 

[pka]

\(\displaystyle \L
\begin{array}{l}
y = \frac{{\left( {x + 3} \right)^4 \left( {2x - 1} \right)^9 }}{{\left( {3x + 1} \right)^3 }} \\
y' = \frac{{\left\{ {4(x + 3)^3 + 18(2x - 1)^8 (x + 3)^4 } \right\}(3x + 1)^3 - \left\{ {3(3x + 1)^2 (3)} \right\}\left( {x + 3} \right)^4 \left( {2x - 1} \right)^9 }}{{(3x + 1)^6 }} \\
\end{array}\)

 

[stapel]

i am this far

(3x+1) [ (x+3)^7 (2x+1)^8 (18) + (2x-1)^9 (x+3)^6 (7)] - (x+3)^7 (2x-1)^9 (9)

I'm not getting the same result at all.

Please reply showing your steps. Thank you.

Eliz.