4 to the right and 3 left
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Answer:
h'(x) = (-3x^3 +2x)/e^(3x)
Step-by-step explanation:
The formula for the derivative of a quotient is ...
(u/v)' = (vu' -uv')/v²
Using u = (x^3 +x^2) and v = e^(3x), this becomes ...
h'(x) = (e^(3x)(3x^2 +2x) -(x^3 +x^2)(3)(e^(3x))/e^(6x)
h'(x) = (-3x^3 +2x)/e^(3x)
Answer: 2/3 * (× + 2 )
Step-by-step explanation:
((x²-4)/(3x)) ÷ ((x-2)/(2x)). ⇒ [ ( ײ - 4 ) * 2x ] ÷ [ ( × - 2 ) *3x ]
Simplifying by x [ 2 * ( ײ - 4 ) ] ÷ [ ( × - 2 ) *3 ] ⇒ (2/3)*{ [ ( x-2 )*(×+2)]÷ (×-2) }
Simplifying by ( ×+2) 2/3 * (× + 2 )