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If we have a function of the form $y=f(g(x))$, sometimes described as a function of a function, to calculate its derivative we need to use the chain rule:
\n\\[ \\frac{dy}{dx} = \\frac{du}{dx} \\times \\frac{dy}{du}.\\]
\n\nThis can be split up into stages:
\nCalculate the derivative of $y=\\simplify[all,fractionNumbers]{{k}({a}*x^{m}+{b})^{n}}$.
\n", "advice": "Following the process of calculating the derivative of a function using the chain rule, we must first identify $g(x)$. Since the function is of the form $y=f(g(x))$, we are looking for the 'inner' function:
\nSo, for $y=\\simplify[all,fractionNumbers]{{k}({a}*x^{m}+{b})^{n}}$, \\[g(x)=\\simplify[all, fractionNumbers, unitFactor]{{a}x^{m}+{b}}.\\]
\nIf we now set $u=g(x)$, we can rewrite $y$ in terms of $u$ such that $y=f(u)$:
\n\\[y=\\simplify[all, fractionNumbers,unitFactor]{{k}u^{n}}.\\]
\nNext, we calculate the two derivatives $\\frac{du}{dx}$ and $\\frac{dy}{du}$:
\n\\[\\frac{du}{dx}=\\simplify[all,fractionNumbers]{{a*m}x^{m-1}}, \\quad \\frac{dy}{du}=\\simplify[all, fractionNumbers, unitFactor]{{k*n}*u^{n-1}}.\\]
\nPlugging these into the chain rule:
\n\\[ \\begin{split} \\frac{dy}{dx} &= \\frac{du}{dx} \\times \\frac{dy}{du}, \\\\&=\\simplify[all,fractionNumbers]{{a*m}x^{m-1}} \\times\\simplify[all, fractionNumbers, unitFactor]{{k*n}u^{n-1}}, \\\\ &=\\simplify[all, fractionNumbers, unitFactor]{{k*n*a*m}*x^{m-1}u^{n-1}}. \\end{split} \\]
\nFinally, we need to express $\\frac{dy}{dx}$ only in terms of $x$, so we must replace the $u$ terms using the initial substitution $u=\\simplify[all, fractionNumbers, unitFactor]{{a}x^{m}+{b}}$:
\n\\[ \\frac{dy}{dx} = \\simplify[all, fractionNumbers, unitFactor]{{k*n*a*m}*x^{m-1}({a}x^{m}+{b})^{n-1}}.\\]
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