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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The derivative of the function y = x^(2a+3b) / x^(a-b) simplifies to y = x^(a+4b). Consequently, the derivative dy/dx is calculated as (a+4b)x^(a+4b-1). This conclusion is reached through the application of the quotient rule and simplification of exponents, confirming that dy/dx = (a+4b)x^(a+4b-1) holds true for integers a and b.

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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



thx
 
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If you have
[tex] y=\frac{fx}{gx}[/tex]
then
[tex] \frac{dy}{dx}=\frac{gx*(fx)'-fx*(gx)'}{(gx)^2}[/tex]
 
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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