Given the derivative below find the requested higher order derivative. Given, f'(x) = 2x^{7/5},...

Question:

Given the derivative below find the requested higher order derivative. Given, {eq}f'(x) = 2x^{\frac{\displaystyle 7}{\displaystyle 5}} {/eq}, find {eq}f'''(x) {/eq}.

(2) A population of {eq}690 {/eq} bacteria is introduced into a culture and grows in number according to the equation {eq}P(t) = 690(1 + \frac{\displaystyle 6t}{\displaystyle 68+t^2}) {/eq} where {eq}t {/eq} is measured in hours. Find the rate at which the population is growing when {eq}t = 7 {/eq}. Round your answer to two decimal places.

Derivatives:

The power rule is a widely used rule in calculus for taking derivatives of functions in the form of polynomials. The rule states that: {eq}\frac {d}{dx} x^n = n \ x^{n-1} {/eq} This applies in all cases of the power n being a real number.

Answer and Explanation:

(1) We use the power rule: {eq}\displaystyle \frac {d}{dx} x^n = n \ x^{n-1} {/eq} on the function given.

{eq}f'(x)=2x^{7/5} {/eq}

We differentiate with respect to x:

{eq}f''(x) = \displaystyle 2(7/5)x^{2/5} = \frac{14}{5} x^\frac{2}{5} {/eq}

And again take derivative:

{eq}f'''(x) = \displaystyle \frac{14(2)}{5(5)} x^{-\frac{3}{5}}= \frac{28}{25} x^{-\frac{3}{5}} {/eq}

Answer:

{eq}\boxed {f'''(x) = \displaystyle \frac{28}{25} x^{-\frac{3}{5}}} {/eq}

(2)

Bacteria grows in time as {eq}P(t) = \displaystyle 690(1 + \frac{6t}{ 68+t^2}) {/eq}

Find

{eq}dP(t)/dt= \displaystyle 690(\frac{6}{ 68+t^2} + \frac{ 6t (-2t)}{ (68+t^2)^2}) = 4140(\frac{68+t^2-2t^2}{ (68+t^2)^2}) = 4140{\frac{(68-t^2)}{ (68+t^2)^2}} {/eq}

at time = 7 hours:

{eq}P'(7) = \displaystyle 4140{\frac{68-49}{ (68+49)^2}} = 5.746 {/eq}

rate at which the population is growing at 7 hours is: 5.75 bacteria/hour


Learn more about this topic:

Loading...
Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
54K

Related to this Question

Explore our homework questions and answers library