\\[\\simplify[std]{f(x) = ({a} * x^{m}+{c}x^2+{b})^{n}}\\]
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
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\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
Differentiate the following function $f(x)$ using the chain rule.
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The chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
For this example, we let $u=\\simplify[std]{{a} * x^{m}+{c}*x^2+{b}}$ and we have $f(u)=\\simplify[std]{u^{n}}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{m*a}x ^ {m -1}+{2*c}x}\\\\\n \n \\frac{df(u)}{du} &=& \\simplify[std]{{n}u^{n-1}} \\end{eqnarray*}\\]
Hence on substituting into the chain rule above we get:
\n \n \n \n\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{({m*a}x ^ {m-1}+{2*c}x) * ({n}*u^{n-1})}\\\\\n \n &=&\\simplify[std]{{n}*({m*a}x^{m-1}+{2*c}*x)u^{n-1}}\\\\\n \n &=& \\simplify[std]{{n}*({m*a}x^{m-1}+{2*c}*x)({a}*x^{m}+{c}*x^2+{b})^{n-1}}\n \n \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x^{m}+{c}x^2+{b}}$.