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If $y=f(g(x))$ to find $\\frac{dy}{dx}$ , we need to do two things::
\n\\[ \\large \\frac{dy}{dx}=\\frac{du}{dx} \\times \\frac{dy}{du} \\]
\n\nNow it is time to get out your paper and pencil and try similar questions without the help. Carry out the same steps, lay it out the same way but you only need to input the final answer.
", "advice": "We are asked to differentiate:
\n\\[ y=(\\var{xCF}x^{\\var{xP}}-\\var{C})^{\\var{P}} \\]
\n\nRecognising that this is a \"function of another function\", we need to identify the \"innermost\" of the two functions that are involved and substitute $u$
\n\nLet $u=\\var{xCF}x^{\\var{xP}}-\\var{C}$ then $y=u^{\\var{P}}$
\n\nNow, we need to use the approriate techniques to differentiate each of these, for both of these we only need the Power Rule:
\n\nApplying this method gives us:
\n$\\large \\frac{du}{dx}=\\simplify{{xCF2}x^{{xP2}}}$ and $ \\large \\frac{dy}{du}= \\simplify{{P}u^{{P2}}}$
\n\n\n
We now use the Chain Rule formula:
\n$ \\large \\large \\frac{dy}{dx}=\\frac{du}{dx} \\times \\frac{dy}{du} $
\nMake the appropriate substitutions into the formula:
\n\n$ \\large \\frac{dy}{dx}= \\simplify{{xCF2}x^{{xP2}}} \\times \\simplify{{P}u^{{P2}}} $
\n\n\n
Which simplifies to:
\n$ \\large \\frac{dy}{dx}=\\simplify{ ({xCF2}x^{xP2})*({P}u^{P2}) }$
\n\nNow, finally, we must remember that $u$ was a variable that we introduced and was not part of the original problem.
\n\nReplace $u$ from our original substitution to give the final answer:
\n\n$\\large \\frac{dy}{dx}=\\simplify{({xCF2}x^{xP2})*({P}({xCF}x^{xP}-{C})^{P2})}$
\n\n\n\n\n\n\n
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\ndifferentiated bracket power
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