How do you find the derivative of #y = (x^2 + sqrtx + 1 ) / x^(3/2)#?

Question

Emily Ammerman · Accepted Answer

#y' = 1/(2x^(1/2)) - 1/x^2 - 3/(2x^(5/2))#

Simplify first:

#y = x^(2 - 3/2) + x^(1/2- 3/2) + x^(0 - 3/2)#

#y = x^(1/2) + x^(-2/2) + x^(-3/2)#

#y = x^(1/2) + x^(-1) + x^(-3/2)#

#y' = 1/2x^(-1/2) - x^(-2) -3/2x^(-5/2)#

#y' = 1/(2x^(1/2)) - 1/x^2 - 3/(2x^(5/2))#

I hope this is useful!

Paisley Johnson · Answer

undefined To find the derivative of $ y = \frac{x^2 + \sqrt{x} + 1}{x^{3/2}} $, you can use the quotient rule. The quotient rule states that if $ y = \frac{u}{v} $, where $ u $ and $ v $ are functions of $ x $, then the derivative of $ y $ with respect to $ x $ is given by:

$$ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} $$

First, identify $ u $ and $ v $ in the given function:

$ u = x^2 + \sqrt{x} + 1 $

$ v = x^{3/2} $

Now, find the derivatives of $ u $ and $ v $ with respect to $ x $:

$$ \frac{du}{dx} = 2x + \frac{1}{2\sqrt{x}} $$

$$ \frac{dv}{dx} = \frac{3}{2}x^{1/2} $$

Now, substitute these derivatives into the quotient rule formula:

$$ \frac{dy}{dx} = \frac{x^{3/2} (2x + \frac{1}{2\sqrt{x}}) - (x^2 + \sqrt{x} + 1) \frac{3}{2}x^{1/2}}{(x^{3/2})^2} $$

$$ \frac{dy}{dx} = \frac{2x^{5/2} + \frac{1}{2} - \frac{3x^{3/2}}{2} - \frac{3\sqrt{x}}{2} - \frac{3}{2}x^{1/2}}{x^3} $$

$$ \frac{dy}{dx} = \frac{2x^{5/2} - \frac{3x^{3/2}}{2} - \frac{3\sqrt{x}}{2} + \frac{1}{2}}{x^3} $$

This expression is the derivative of $ y $ with respect to $ x $.