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Calculating the derivative of a function of the form $\\sin(ax^m+bx^n)$ using the chain rule.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Calculate the derivative of $y=\\simplify[all]{sin({a}*x^{n}+{b}*x^{m})}$.
", "advice": "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 steps:
\nFollowing this process, 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 $y=\\simplify[all,fractionNumbers]{sin({a}*x^{n}+{b}*x^{m})}$, \\[g(x)=\\simplify[all, fractionNumbers, unitFactor]{{a}*x^{n}+{b}*x^{m}}.\\]
\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]{sin(u)}.\\]
\nNext, we calculate the two derivatives $\\frac{du}{dx}$ and $\\frac{dy}{du}$:
\n\\[\\frac{du}{dx}=\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}}, \\quad \\frac{dy}{du}=\\simplify[all, fractionNumbers, unitFactor]{cos(u)}.\\]
\nPlugging these into the chain rule:
\n\\[ \\begin{split} \\frac{dy}{dx} &= \\frac{du}{dx} \\times \\frac{dy}{du}, \\\\&=(\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}}) \\times\\simplify[all, fractionNumbers, unitFactor]{cos(u)}. \\end{split} \\]
\nFinally, we need to express $\\frac{dy}{dx}$ only in terms of $x$, so we must replace the $u$ term using the initial substitution $u=\\simplify[all, fractionNumbers, unitFactor]{{a}*x^{n}+{b}*x^{m}}$:
\n\\[ \\frac{dy}{dx} =(\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}})\\simplify[all, fractionNumbers, unitFactor]{cos({a}*x^{n}+{b}*x^{m})}.\\]
\n\nUse this link to find some resources which will help you revise this topic.
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