Answer:
![(dy)/(dx)=\text{ 5\lparen10x - 8\rparen\lparen5x}^2\text{ - 8x + 3\rparen}^4](https://img.qammunity.org/2023/formulas/mathematics/college/xr317b9z6g4k9pjmy4dsde4e3pcmf19xx8.png)
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Step-by-step explanation:
Given:
![y\text{ = \lparen5x}^2-\text{ 8x + 3\rparen}^5](https://img.qammunity.org/2023/formulas/mathematics/college/wfoejwjn2ygwb3lwbeywgl9ru2wrxr94qx.png)
To find:
derivative of the function = dy/dx
To determine the derivative, we will follow the steps provided in the question
![\begin{gathered} a)\text{ }let\text{ the inside = expression in the parenthesis} \\ \text{inside: u = 5x}^2\text{ - 8x + 3} \\ \\ Substitute\text{ for u} \\ y\text{ = u}^5 \\ outside:\text{ y = u}^5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fytw6ka93oynfgfejm4op546z6je38ybvs.png)
b) Next is to find the derivative of the outside and insides:
![\begin{gathered} inside:\text{ u = 5x}^2\text{ - 8x + 3} \\ (du)/(dx)\text{ = 10x - 8} \\ \\ outside:\text{ y = u}^5 \\ (dy)/(du)\text{ = 5u}^(5-1) \\ (dy)/(du)\text{ = 5u}^4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lgivolaxexapwp3v7jfcjy6eekd835vdbe.png)
c) We will use the chain rule formula to get the derivative:
![\begin{gathered} (dy)/(dx)=\text{ }(dy)/(du)*(du)/(dx) \\ \\ (dy)/(dx)=\text{ 5u}^4\text{ }*\text{ \lparen10x - 8\rparen} \\ \\ recall\text{ u = 5x}^2\text{ - 8x + 3} \\ substitute\text{ for u in the derivative above:} \\ (dy)/(dx)=\text{ 5\lparen5x}^2-\text{ 8x + 3\rparen}^4\text{ }*\text{ \lparen10x - 8\rparen} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gnqcbnxaqg3xxmtkneaqwsbwviqiul8449.png)
![(dy)/(dx)=\text{ 5\lparen10x - 8\rparen\lparen5x}^2\text{ - 8x + 3\rparen}^4](https://img.qammunity.org/2023/formulas/mathematics/college/xr317b9z6g4k9pjmy4dsde4e3pcmf19xx8.png)