Proving dy/dx = (a+4b)x^(a+4b-1) for y = x^(2a+3b)/x^(a-b), a and b Integers

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To prove that dy/dx = (a+4b)x^(a+4b-1) for y = x^(2a+3b)/x^(a-b), the expression can be simplified to y = x^(a+4b). The derivative of y, denoted as y', is calculated using the power rule. Applying the power rule yields y' = (a+4b)x^(a+4b-1). Thus, the proof confirms that dy/dx indeed equals (a+4b)x^(a+4b-1).
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Show that dy/dx = (a+4b)x^(a+4b-1) if y = x^(2a+3b) / x^(a-b) and a and b are integers



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If you have
<br /> y=\frac{fx}{gx}<br />
then
<br /> \frac{dy}{dx}=\frac{gx*(fx)&#039;-fx*(gx)&#039;}{(gx)^2}<br />
 
Much more simply:
y = x^(2a+3b) / x^(a-b)
= x^((2a+3b)-(a-b))
= x^(a+ 4b)

Now, what is y'?
 

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