# differentitation help please !!!!

• Jun 24th 2009, 12:22 PM
lukyleo26
differentitation help please !!!!
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, 12:32 PM
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, 12:38 PM
Random Variable
a) let $u = x^{3} + 5$ and use the chain rule

b) product rule

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

c) quotient rule

$\frac{(x+3)\frac{d}{dx}e^{2x} - e^{2x}\frac {d}{dx} (x+3)}{(x+3)^2}$
• Jun 24th 2009, 12:39 PM
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)

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

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

Quote:

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

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

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

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

Quote:

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

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

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

$y' = \frac{2(x+3)e^{2x} - e^{2x}}{(x+3)^2} = \frac{e^{2x}(2x+5)}{(x+3)^2}$
• Jun 24th 2009, 12:40 PM
lukyleo26
for b is the answer

yx^3*sin2*X

Fx=sin2*yx^3

f/x= sin2yx^3