1) You need to differentiate the function `y = (3x-4)/(x^3)` with respect to x.

Since the x variable is present both numerator and denominator, you need to use the quotient rule when differentiate:

`dy/dx = ((3x-4)'*(x^3) - (3x-4)*(x^3)')/((x^3)^2)`

`dy/dx = (3x^3 - (3x^2)*(3x-4))/(x^6)`

`` `dy/dx = (3x^3 - 9x^3 + 12x^2)/(x^6)`

`dy/dx = (- 6x^3 + 12x^2)/(x^6)`

Factorizing by `6x^2` yields:

`dy/dx = (6x^2)(- x + 2)/(x^6) =gt dy/dx = (12 - 6x)/(x^4)`

`` 2) You need to use the product rule to differentiate `y = x sqrt(x^2+5).`

`` `dy/dx = (x)'* sqrt(x^2+5) + x *(sqrt(x^2+5))'`

`dy/dx = sqrt(x^2+5) + (x *2x)/(2sqrt(x^2+5))`

`dy/dx = sqrt(x^2+5) + (2x^2)/(2sqrt(x^2+5))`

You need to bring both fractions to a common denominator:

`dy/dx = (x^2 + 5 + 2x^2)/(2sqrt(x^2+5))`

`` `dy/dx = (3x^2 + 5)/(2sqrt(x^2+5))`

**Differentiating both functions, using the quotient and the product rules, yields: `dy/dx = (12 - 6x)/(x^4)` ; `dy/dx = (3x^2 + 5)/(2sqrt(x^2+5)).` **

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