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

By rewriting a bit,

By Quotient Rule,

By cleaning up a bit,

I hope that this was clear.

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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).

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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} ]

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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.

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.

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.

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