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Final Answer

$\frac{\left(16\left(2ax+a^2\right)-25\right)\left(16a^3-7a^2-10a\right)+\left(-16a^2x+25x\right)\left(48a^{2}-14a-10\right)}{\left(16a^3-7a^2-10a\right)^2}$
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Step-by-step Solution

1

Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\frac{d}{dx}\left(16a^2x-25x\right)\left(16a^3-7a^2-10a\right)-\left(16a^2x-25x\right)\frac{d}{dx}\left(16a^3-7a^2-10a\right)}{\left(16a^3-7a^2-10a\right)^2}$

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$\frac{\frac{d}{dx}\left(16a^2x-25x\right)\left(16a^3-7a^2-10a\right)-\left(16a^2x-25x\right)\frac{d}{dx}\left(16a^3-7a^2-10a\right)}{\left(16a^3-7a^2-10a\right)^2}$

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Learn how to solve problems step by step online. . Apply the quotient rule for differentiation, which states that if f(x) and g(x) are functions and h(x) is the function defined by {\displaystyle h(x) = \frac{f(x)}{g(x)}}, where {g(x) \neq 0}, then {\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}. Simplify the product -(16a^2x-25x). The derivative of a sum of two or more functions is the sum of the derivatives of each function. The derivative of a sum of two or more functions is the sum of the derivatives of each function.

Final Answer

$\frac{\left(16\left(2ax+a^2\right)-25\right)\left(16a^3-7a^2-10a\right)+\left(-16a^2x+25x\right)\left(48a^{2}-14a-10\right)}{\left(16a^3-7a^2-10a\right)^2}$

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