# rate of change dy/dx

• Jun 20th 2006, 07:53 PM
becky
rate of change dy/dx
Find the rate of change (dy/dx) for the prescribed value of x.

y = (x^2 + 2) (x + square root of x) x=4
y = (x^2 + 2)(x + x^-1/2)
y = (x^2 + 2)(1/2square root x) + (x + 1/2square root of x)(2x)

If that is correct thus far, how do I go about multipling this out?
• Jun 20th 2006, 08:35 PM
earboth
Quote:

Originally Posted by becky
Find the rate of change (dy/dx) for the prescribed value of x.
y = (x^2 + 2) (x + square root of x) x=4
y = (x^2 + 2)(x + x^-1/2)
y = (x^2 + 2)(1/2square root x) + (x + 1/2square root of x)(2x)
If that is correct thus far, how do I go about multipling this out?

Hello,

you've got:

$\displaystyle y=(x^2+2)(x+x^{\frac{1}{2}})$

Obviously you don't know the product rule to calculate the derivative of y. So you have to multiply first:

$\displaystyle y=(x^2+2)(x+x^{\frac{1}{2}})=x^3+x^{\frac{5}{2}}+2 x+2x^{\frac{1}{2}}$

Now differentiate:

$\displaystyle \frac{dy}{dc}=3x^2+{5\over 2}x^{\frac{3}{2}}+2+ x^{-\frac{1}{2}}$

Plug in the value you know and you'll get:

$\displaystyle 3(4)^2+{5\over 2}(4)^{\frac{3}{2}}+2+ (4)^{-\frac{1}{2}}$ =

$\displaystyle 48+20+2+ \frac{1}{2}} = \frac{141}{2}$

Greetings

EB