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scoundrel [369]
3 years ago
15

Find f'(x) for the given function. f(x) = 5/3-x

Mathematics
1 answer:
Wewaii [24]3 years ago
8 0

Answer:

The derivative is:

f^{\prime}(x) = -\frac{5}{x^2-6x+9}

Step-by-step explanation:

We are given the following function:

f(x) = \frac{5}{3-x}

Derivative of a quotient:

Suppose we have a quotient:

f(x) = \frac{g(x)}{h(x)}

The derivative is:

f^{\prime}(x) = \frac{g^{\prime}(x)h(x) - h^{\prime}(x)g(x)}{h(x)^2}

In this question:

Numerator g(x) = 5, g^{\prime}(x) = 0

Denominator h(x) = 3 - x, h^{\prime}(x) = -1

So

f^{\prime}(x) = \frac{0(3-x) - 5}{(3 - x)^2} = -\frac{5}{x^2-6x+9}

The derivative is:

f^{\prime}(x) = -\frac{5}{x^2-6x+9}

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