1. ## 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 !!!!!

2. 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.

3. 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

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

4. 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:
$y' = \frac{3}{2} \cdot \frac{x^2}{\sqrt{x^3+5}}$

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

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

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

$y' = \frac{2(x+3)e^{2x} - e^{2x}}{(x+3)^2} = \frac{e^{2x}(2x+5)}{(x+3)^2}$

5. for b is the answer

yx^3*sin2*X

Fx=sin2*yx^3

f/x= sin2yx^3