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Sophie [7]
3 years ago
6

Derivative calculus

Mathematics
1 answer:
mestny [16]3 years ago
3 0

Answer:

\mathsf{ \blue{ f'(x) =  \frac{ - 10}{ {(x - 5)}^{2} } } }

Step-by-step explanation:

\mathsf{f(x) =  \frac{ {x}^{2}  + 10x + 25}{ {x}^{2}  - 25} }

the above expression can be reduced to simpler terms

  • x² - 25 = (x + 5)(x - 5)
  • x² + 10x + 25 = (x + 5)²

\mathsf{\implies f(x) = \frac{ {(x + 5)}^{2} }{(x + 5)(x - 5)}  }

(x + 5)² can be written as (x + 5)(x + 5)

\mathsf{\implies f(x) = \frac{ \cancel{(x + 5)}(x + 5)}{\cancel{(x + 5)}(x - 5)}  }

\mathsf{\implies f(x) = \frac{x + 5}{x - 5}  }

Derivative of a fraction \mathsf{\frac{u}{v}} is

  • \boxed { \red {\mathsf{ \frac{v(\frac{du}{dx}) \:  - \: u(\frac{dv}{dx})}{v^2}} }}

\mathsf{\implies f'(x) =  \frac{(x  -  5) - (x  +  5)}{(x - 5) {}^{2} } }

\mathsf{\implies f '(x) =   \frac{x  -  5 - x  -  5}{ {(x - 5)}^{2} }  }

\mathsf{\implies f'(x) =  \frac{ - 10}{ {(x - 5)}^{2} } }

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