# Differentiation questions.

• Oct 3rd 2010, 10:45 AM
daniel123
I am also on the next question, i have solved a few and struggling with others?

2. obtain dy/dx for the following functions using the 'direct rule'

a) y=2.5x^2
=5x

b) y=x^2/2.5
=0.8x

c) y=2.5x - 2.5
=-2.5x^0

d)y= (2.5)(SQUAREROOT OF x)
=1.25x

e)y=x^2.5
=2.5^1.5

f)y=2.5/x^2 HELP

g)y=(^2.5 SQUARE ROOT OF x) HELP

h)y=2.5x^4 - (x^3/2.5) - (x/2.5) + 2.5 SO FAR
=10x^3 - 1.2x^2 - 0.4x^0 + 2.5

I)y=2.5x{(x^2)-(1)} HELP

J) y=(x^3 - 2.5x^2) / (2.5) (SQUARE ROOT OF x) HELP
• Oct 3rd 2010, 12:48 PM
lvleph
d)
$\displaystyle y = 2.5 \sqrt(x) = 2.5 x^{0.5}$
By the power rule
$\displaystyle y' = 0.5*2.5 x^{0.5-1} = \dfrac{1.25}{\sqrt{x}}$

f)
$\displaystyle y = \dfrac{2.5}{x^2} = 2.5 x^{-2}$ now use the power rule. [Typo Fixed]

g) What?

h) If C is a constant then $\displaystyle \frac{d}{dx}[C] = 0$
I)
$\displaystyle y=2.5 x(x^2 -1) = 2.5x^3 - 2.5x$ Now use a combination of power rule and sum rule

J)
$\displaystyle y=\dfrac{x^3 - 2.5x^2}{2.5\sqrt{ x}} = 0.4x^{2.5} - x^{1.5}$ Now use a combination of the power rule and sum rule.
• Oct 5th 2010, 03:50 PM
daniel123
Ok i have attempted the questions you have helped me with:

d) 1.25^-0.5x

f) 5x

h) 10x^3 - 1.2x - 0.4x + 2.5

I) (7.5x^2) - (2.5x)

J) (x^1.5) - (1.5x^0.5)

Thankyou
• Oct 5th 2010, 07:24 PM
Educated

e) $\displaystyle y=x^{2.5}$
$\displaystyle \frac{dy}{dx}=2.5^{1.5}$

Where'd your x go? If it's $\displaystyle \frac{dy}{dx}=2.5x^{1.5}$ then it is correct.

f) y=2.5/x^2
Another way of writing this is $\displaystyle y=2.5x^{-2}$

g) Doesn't make sense. Is it $\displaystyle y= \sqrt{x}^{2.5})$??

h) $\displaystyle y=2.5x^4 - \frac{x^3}{2.5} - \frac{x}{2.5} + 2.5$
You are almost there. Why do you still have that + 2.5 part at the end?

i) (7.5x^2) - (2.5x)
Almost there with this one as well. You haven't seemed to have differentiated the 2.5x part yet.

j) Correct.
• Oct 6th 2010, 03:58 AM
HallsofIvy
Quote:

Originally Posted by lvleph
d)
$\displaystyle y = 2.5 \sqrt(x) = 2.5 x^{0.5}$
By the power rule
$\displaystyle y' = 0.5*2.5 x^{0.5-1} = \dfrac{1.25}{\sqrt{x}}$

f)
$\displaystyle y = \dfrac{2.5}{x^2} = 2.5 x^2$ now use the power rule.

Typo! lvleph surely meant $\displaystyle 2.5 x^{-2}$

Quote:

g) What?

h) If C is a constant then $\displaystyle \frac{d}{dx}[C] = 0$
I)
$\displaystyle y=2.5 x(x^2 -1) = 2.5x^3 - 2.5x$ Now use a combination of power rule and sum rule

J)
$\displaystyle y=\dfrac{x^3 - 2.5x^2}{2.5\sqrt{ x}} = 0.4x^{2.5} - x^{1.5}$ Now use a combination of the power rule and sum rule.
• Oct 6th 2010, 05:48 AM
lvleph
Maybe g) is $\displaystyle x^{1/2.5}$?