What is the derivative of #y= (5x)/sqrt (x^2+9)#?

Question

William Abdalla · Accepted Answer

#y'=45/(x^2+9)^(3/2)#

By rewriting a bit,

#y=5 x/(x^2+9)^(1/2)#

By Quotient Rule,

#y'=5(1 cdot (x^2+9)^(1/2)-x cdot1/(cancel2)(x^2+9)^(-1/2)(cancel 2x))/((x^2+9)^(1/2))^2#

By cleaning up a bit,

#=5((x^2+9)^(1/2)-x^2/(x^2+9)^(1/2))/(x^2+9)#

By multiply the numerator and the denominator by #(x^2+9)^(1/2)#,

#=5(cancel(x^2)+9-cancel(x^2))/(x^2+9)^(3/2)=45/(x^2+9)^(3/2)#

I hope that this was clear.

Charles Arocha · Answer

undefined To find the derivative of the function y = (5x)/sqrt(x^2 + 9), you can use the quotient rule. The quotient rule states that if you have a function in the form of u/v, where u and v are both functions of x, the derivative is (v * u' - u * v') / v^2, where u' and v' are the derivatives of u and v, respectively. Applying this rule to the given function:

Let u = 5x and v = sqrt(x^2 + 9).

Then, u' = 5 and v' = (1/2)*(x^2 + 9)^(-1/2)*(2x) = x/(sqrt(x^2 + 9)).

Now, applying the quotient rule:

y' = (v * u' - u * v') / v^2
   = (sqrt(x^2 + 9) * 5 - 5x * (x/(sqrt(x^2 + 9)))) / (sqrt(x^2 + 9))^2
   = (5sqrt(x^2 + 9) - 5x^2/(sqrt(x^2 + 9))) / (x^2 + 9).

Thus, the derivative of y = (5x)/sqrt(x^2 + 9) is:

y' = (5sqrt(x^2 + 9) - 5x^2/(sqrt(x^2 + 9))) / (x^2 + 9).

Luke Alvarez · Answer

undefined The derivative of $ y = \frac{5x}{\sqrt{x^2 + 9}} $ with respect to $ x $ can be found using the quotient rule. The derivative is:

$$ \frac{d}{dx} \left( \frac{5x}{\sqrt{x^2 + 9}} \right) = \frac{5(x^2 + 9)^{\frac{1}{2}} - 5x \cdot \frac{1}{2}(x^2 + 9)^{-\frac{1}{2}}(2x)}{x^2 + 9} $$