How do you differentiate #y= (sqrt(x)+x) / x^2#?

Question

Isabella Ackerman · Accepted Answer

#-(3/(2x^(5/2)) +1/x^2)#

divide terms on numerator by #x^2#

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

now differentiate each term using the #color(blue)" power rule "#

#dy/dx=-3/2 x^(-5/2)-1x^-2#

and writing with positive indices gives.

#dy/dx=-(3/(2x^(5/2))+1/x^2)#

Scarlett Allison · Answer

undefined To differentiate the function $ y = \frac{\sqrt{x} + x}{x^2} $, you can use the quotient rule. The quotient rule states that for a function $ u(x) $ divided by $ v(x) $, the derivative is $ \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $. Applying this rule to the given function:

Let $ u(x) = \sqrt{x} + x $ and $ v(x) = x^2 $.

Now, differentiate both $ u(x) $ and $ v(x) $ separately:

$ u'(x) = \frac{d}{dx}(\sqrt{x} + x) = \frac{1}{2\sqrt{x}} + 1 $

$ v'(x) = \frac{d}{dx}(x^2) = 2x $

Now, apply the quotient rule:

$ y' = \frac{(1/2\sqrt{x} + 1)(x^2) - (\sqrt{x} + x)(2x)}{(x^2)^2} $

$ y' = \frac{(x^2/2\sqrt{x}) + x^2 - 2x\sqrt{x} - 2x^2}{x^4} $

$ y' = \frac{x^{5/2} + x^2 - 2x\sqrt{x} - 2x^2}{2x^4} $

So, the derivative of $ y = \frac{\sqrt{x} + x}{x^2} $ is $ y' = \frac{x^{5/2} + x^2 - 2x\sqrt{x} - 2x^2}{2x^4} $.