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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\nWe are asked to differentiate:
\n\\[ y=\\ln({\\simplify{{CF}x}+\\sin{(x)})} \\]
\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 $u=\\simplify{{CF}x+sin(x)}$ then $y=\\ln(u)$
\n\nNow, we need to use the approriate techniques to differentiate each of these, for these functions we need the Power Rule and your Table of Derivatives.:
\n\nApplying this method gives us:
\n$\\large \\frac{du}{dx}=\\var{CF}+cos(x)$ and $ \\large \\frac{dy}{du}= \\frac{1}{u}$
\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}= (\\var{CF}+cos(x)) \\times \\frac{1}{u}$
\n\n\n
Which simplifies to:
\n$ \\large \\frac{dy}{dx}=\\frac{\\var{CF}+cos(x)}{u}$
\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$\\large \\frac{dy}{dx}=\\frac{\\var{CF}+cos(x)}{\\simplify{{CF}x+sin(x)}}$
\n\n\n\n\n\n\n
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First identify the \"innermost\" function, and substitute $u$:
\nLet $u=$[[0]] Then $y=$[[1]]
\nThen:
\n$ \\large \\frac{du}{dx}= $[[2]] and $ \\large \\frac{dy}{du}= $[[3]]
\nNow using:
\n$\\large \\frac{dy}{dx}=\\frac{du}{dx} \\times \\frac{dy}{du}$
\n\n$\\large \\frac{dy}{dx}=$[[4]]$\\times$[[5]]
\nWhich simplifies to:
\n$\\large \\frac{dy}{dx}=$[[6]]
\n\nRemember that $u$ was a variable that we introduced and not part of the original problem.
\nReplace $u$ from our substitution to give the final answer:
\n$\\large \\frac{dy}{dx}=$[[7]]
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