1. ## help differentiation

a)...A rectangular page is to contain 18 square inches of print. The
margins at the top and bottom are .5 inches and on each side are 1 inch. What dimensions minimize the amount of paper used?

my answer was 8 x 4 dimention.. is it correct?
b)...Differentiate the function f(t) = (2x-5)e^3x-1
i cant solve it please if you could help in that.???

2. ## Re: help differentiation

I'm getting a length of 4 and a width of 8. I don't know if this matches your 8x4. Usually length is specified first. I'm assuming that length corresponds to the measure from the top to the bottom.

for (b) just use the product rule

$\dfrac d {dt} (2t-5)e^{3t-1} = \left(\dfrac{d}{dt}(2t-5)\right) \cdot e^{3t-1} + (2t-5)\cdot \left(\dfrac{d}{dt}e^{3t-1}\right)$

3. ## Re: help differentiation

Noname.bmp

is it correct??

4. ## Re: help differentiation

Originally Posted by rabia123

Noname.bmp

is it correct??
I do not get your answer, mostly because you forgot the multiplication rule, which rule you could have avoided by using logs more imaginatively. This is an ugly problem so please check my algebra for mistakes.

$y = \dfrac{(4x^2 - 5)^2}{(2x - 3)^4(5x^4 - 2)^5(3x^2 - 4)^3} \implies ln(y) = ln\left(\dfrac{(4x^2 - 5)^2}{(2x - 3)^4(5x^4 - 2)^5(3x^2 - 4)^3}\right) \implies$

$ln(y) = ln\left\{(4x^2 - 5)^2\right\} - \left(ln\left\{(2x - 3)^4(5x^4 - 2)^5(3x^2 - 4)^3\right\}\right) =$

$ln\left\{(4x^2 - 5)^2\right\} - \left(ln\left\{(2x - 3)^4\right\} + ln\left\{(5x^4 - 2)^5\right\} + ln\left\{(3x^2 - 4)^3\right\}\right) =$

$ln\left\{(4x^2 - 5)^2\right\} - \left(ln\left\{(2x - 3)^4\right\} + ln\left\{(5x^4 - 2)^5\right\} + ln\left\{(3x^2 - 4)^3\right\}\right) \implies$

$ln(y) = 2ln(4x^2 - 5) - 4ln(2x - 3) - 5ln(5x^4 - 2) - 3ln(3x^2 - 4) \implies$

$\dfrac{1}{y} * \dfrac{dy}{dx} = \dfrac{2 * 8x}{4x^2 - 5} - \dfrac{4 * 2}{2x - 3} - \dfrac{5 * 20x^3}{5x^4 - 2} - \dfrac{3 * 6x}{3x^2 - 4} =$

$\dfrac{16x}{4x^2 - 5} - \dfrac{8}{2x - 3} - \dfrac{100x^3}{5x^4 - 2} - \dfrac{18x}{3x^2 - 4} \implies$

$\dfrac{dy}{dx} = y\left(\dfrac{16x}{4x^2 - 5} - \dfrac{8}{2x - 3} - \dfrac{100x^3}{5x^4 - 2} - \dfrac{18x}{3x^2 - 4}\right) =$

$\left(\dfrac{(4x^2 - 5)^2}{(2x - 3)^4(5x^4 - 2)^5(3x^2 - 4)^3}\right)\left(\dfrac{16x}{4x^2 - 5} - \dfrac{8}{2x - 3} - \dfrac{100x^3}{5x^4 - 2} - \dfrac{18x}{3x^2 - 4}\right) \implies$

$\dfrac{dy}{dx} = \dfrac{16x(4x^2 - 5)}{(2x - 3)^4(5x^4 - 2)^5(3x^2 - 4)^3} - \dfrac{8(4x^2 - 5)^2}{(2x - 3)^5(5x^4 - 2)^5(3x^2 - 4)^3} - \dfrac{100x^3(4x^2 - 5)^2}{(2x - 3)^4(5x^4 - 2)^6(3x^2 - 4)^3} - \dfrac{18x(4x^2 - 5)^2}{(2x - 3)^4(5x^4 - 2)^5(3x^2 - 4)^4}.$

Again, I have tried to avoid algebraic mistakes, but do not guarantee I have found them all. I dislike this problem because it involves tedious and therefore error provoking algebra. The fundamental point, however, is that you either have to use the multiplication rule in the differentiation or complete the transition to logs. That is the calculus part of the problem.

5. ## Re: help differentiation

Just to highlight this principle as it's not completely clear from JeffM's work.

$\dfrac d {dx} \ln(f(x))=\dfrac {f^\prime(x)}{f(x)}$ so

$f^\prime(x)=f(x) \dfrac d {dx} \left(\ln(f(x))\right)$