Find dp/dq If P= q²+3/(q-1)³+(q+1)³
"p=q\u00b2+\\frac{3}{(q-1)\u00b3}+(q+1)\u00b3"
Differentiate both sides of the equation
"\\frac{dp}{dq}=\\frac{d}{dq}(q\u00b2+\\frac{3}{(q-1)\u00b3}+(q+3)\u00b3"
The derivative of p with respect to q is
"p'" .
By the sum rule,
"\\implies p'=\\frac{d}{dq}(q\u00b2)+\\frac{d}{dq}\\frac{3}{(q-1)\u00b3}+\\frac{d}{dq}(q+1)\u00b3"
Differentiating the right side of the equation
"p'=2q-\\frac{9}{(q-1)\u2074}+3(q+1)\u00b2"
"\\implies \\frac{dp}{dq}=2q-\\frac{9}{(q-1)\u2074}+3(q+1)\u00b2"
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