How do you find the derivative with a square root in the denominator #y= 5x/sqrt(x^2+9)#?

Answer 1

#dy/dx=(10(x^2+9)-5x)/(sqrt(x^2+9)(2x^2+18))#

The Quotient Rule says

#d/dx(f(x)/g(x))=(g(x)d/dx(f(x))-f(x)d/dx(g(x)))/(g(x))^2#

If we let #f(x) = 5x# and #g(x)=sqrt(x^2+9)#, so that #f'(x)=5# and #g'(x)=1/2(x^2+9)^(-1/2)#, and #(g(x))^2=x^2+9#. Plugging these results in the Quotient Rule, we then have

#dy/dx=d/dx(f(x)/g(x))=(5sqrt(x^2+9)-5x1/2(x^2+9)^(-1/2))/(x^2+9)# #=((2 xx)/(2 xx))(5sqrt(x^2+9)-5x1/2(x^2+9)^(-1/2))/(x^2+9)# #=((sqrt(x^2+9) xx)/(sqrt(x^2+9) xx))(10sqrt(x^2+9)-5x(x^2+9)^(-1/2))/(2x^2+18)# #=(10(x^2+9)-5x)/(sqrt(x^2+9)(2x^2+18))#

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Answer 2

Ezekiel Alexander

To find the derivative of the function ( y = \frac{5x}{\sqrt{x^2 + 9}} ) with a square root in the denominator, we can use the quotient rule. Let ( u = 5x ) and ( v = \sqrt{x^2 + 9} ). Then, ( u' = 5 ) and ( v' = \frac{x}{\sqrt{x^2 + 9}} ).

Now, applying the quotient rule:

[ y' = \frac{u'v - uv'}{v^2} = \frac{5\sqrt{x^2 + 9} - \frac{5x^2}{\sqrt{x^2 + 9}}}{x^2 + 9} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.