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Chain rule in calculus

User Knodumi
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\begin{gathered} \text{If we have }y=f(u),\text{ and }u=g(x).\text{ Then by chain rule, the derivative of }y\text{ is} \\ (dy)/(dx)=(dy)/(du)\cdot(du)/(dx) \end{gathered}

In the given example:


\begin{gathered} u=4x^3-5 \\ f(u)=u^4 \\ \text{If we do a function composition then they will be the same} \\ f(x)=\big(4x^3-5\big)^4\rightarrow f(u)=u^4,\text{ note that }u=4x^3-5 \end{gathered}

Solve for each derivative of dy/du and du/dx


\begin{gathered} (du)/(dx)=3\cdot4x^(3-1)-0 \\ (du)/(dx)=12x^2 \\ \\ (dy)/(du)=4\cdot u^(4-1) \\ (dy)/(du)=4u^3,\text{ then substitute }u \\ (dy)/(du)=4(4x^3-5)^3 \\ \\ \text{Complete the chain rule} \\ (dy)/(dx)=(dy)/(du)\cdot(du)/(dx) \\ (dy)/(dx)=\big(4(4x^3-5)^3\big)\big(12x^2\big)\text{ or }(dy)/(dx)=48x^2(4x^3-5)^3 \\ \end{gathered}

User Kiril Dobrev
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