• Jun 24th 2009, 11:22 AM
lukyleo26
could anyone help me with these questions please i have a few answers myslef but dont know if i am going the right way with this or not. im useless at differentiation please be kind lol !!!

Q1, differentiate the following with respect to X

a, Y=squareroot of(x^3+5)

p.s dnt know how to get square root sign on p.c

b, Y=X^3sin2x

c, Y=e^2x /(x+3)

thankyou for your time hopefully some of you are better at this than me lol im sure you are !!!!!:)
• Jun 24th 2009, 11:32 AM
HallsofIvy
We can't say if you are "going the right way" if you don't show us which way you are going! Show what you are doing, please.
• Jun 24th 2009, 11:38 AM
Random Variable
a) let $\displaystyle u = x^{3} + 5$ and use the chain rule

b) product rule

$\displaystyle \sin(2x) \frac{d}{dx} x^{3} + x^{3} \frac {d}{dx} \sin(2x)$

c) quotient rule

$\displaystyle \frac{(x+3)\frac{d}{dx}e^{2x} - e^{2x}\frac {d}{dx} (x+3)}{(x+3)^2}$
• Jun 24th 2009, 11:39 AM
e^(i*pi)
Quote:

Originally Posted by lukyleo26
could anyone help me with these questions please i have a few answers myslef but dont know if i am going the right way with this or not. im useless at differentiation please be kind lol !!!

Q1, differentiate the following with respect to X

a, Y=squareroot of(x^3+5)

$\displaystyle y = \sqrt{x^3+5} = (x^3+5)^{\frac{1}{2}}$. You can then use the chain rule to solve

Spoiler:
$\displaystyle y' = \frac{3}{2} \cdot \frac{x^2}{\sqrt{x^3+5}}$

Quote:

b, Y=X^3sin2x
Product Rule: $\displaystyle y=f(x)g(x) \: , \: y' = f'(x)g(x) + g'(x)f(x)$

In your case $\displaystyle f(x) = x^3$ and $\displaystyle g(x) = sin(2x)$. Note that you must use the chain rule on sin(2x).

Spoiler:
$\displaystyle u = x^3 \: , \: v = sin(2x)$
$\displaystyle u' = 3x^2 \: , \: v' = 2cos(2x)$

$\displaystyle y' = 2x^3cos(2x) + 3x^2sin(2x)$

Quote:

c, Y=e^2x /(x+3)
Quotient Rule:$\displaystyle y = \frac{u}{v} \: , \: y' = \frac{vu' - uv'}{v^2}$

In your case $\displaystyle u = e^{2x}$ (and don't forget the chain rule) and $\displaystyle v = x+3$

Spoiler:
$\displaystyle u = e^{2x} \: , \: v = x+3$
$\displaystyle u' = 2e^{2x} \: , \: v' = 1$

$\displaystyle y' = \frac{2(x+3)e^{2x} - e^{2x}}{(x+3)^2} = \frac{e^{2x}(2x+5)}{(x+3)^2}$
• Jun 24th 2009, 11:40 AM
lukyleo26