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A biological model' measures the production of a certain type of white blood cell (granulocytes)...

Question:

A biological model' measures the production of a certain type of white blood cell (granulocytes) by the function

{eq}p(x) = \frac{Ax}{B + x^m} {/eq}

where A and B are positive constants, the exponent m is positive, and x is the number of cells present.

a. Find the rate of production p'(x).

b. Find p"(x), and determine all values of x for which p"(x) = 0 (your answer will involve m).

Derivatives

The derivative of a function, {eq}f(x) {/eq}, measures the rate of change in the function, {eq}f(x) {/eq}, as the variable {eq}x {/eq} is changing. Taking the derivative or differentiation is the reverse process of integration. The derivative of the position function with respect to time is the velocity function.

Answer and Explanation:


The derivative of the function {eq}\displaystyle \frac{f(x)}{g(x)} {/eq} can be calculated using the quotient rule

{eq}\displaystyle \frac{d}{dx}\frac{f(x)}{g(x)} = \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}. {/eq}


a.) Solving for {eq}p'(x) {/eq}.


{eq}\displaystyle \begin{align*} p'(x) &= \frac{d}{dx} p(x)\\ &= \frac{d}{dx} \bigg( \frac{Ax}{B + x^m}\bigg)\\ &= \frac{(B + x^m)\frac{d\ Ax}{dx} - (Ax) \frac{d (B + x^m)}{dx}}{(B + x^m)^2}\\ &= \frac{A(B + x^m) -(Ax) mx^{m-1}}{(B + x^m)^2}\\ &= \frac{A (B + x^m) -(x) mx^{m-1}}{(B + x^m)^2}\\ &=\boxed{\frac{A (B + x^m-mx^m)}{(B + x^m)^2}}\\ \end{align*} {/eq}


b.) Solving for all values of x such that {eq}p''(x) = 0 {/eq}.


We start with solving {eq}p''(x) {/eq}.


{eq}\displaystyle \begin{align*} p''(x) &= \frac{d}{dx}p'(x) \\ &= \frac{d}{dx}\bigg( \frac{A (B + x^m-mx^m)}{(B + x^m)^2}\bigg)\\ &= \frac{(B + x^m)^2\frac{d A (B + x^m-mx^m)}{dx} - [A (B + x^m-mx^m)]\frac{d (B + x^m)^2}{dx}}{[(B + x^m)^2]^2}\\ &= \frac{(B + x^m)^2 (A)(mx^{m-1}-m^2x^{m-1}) - [A (B + x^m-mx^m)](2)mx^{m-1}(B + x^m)^{2-1}}{[(B + x^m)^2]^2}\\ &= \frac{(B + x^m)\bigg[(B + x^m)(A)(mx^{m-1}-m^2x^{m-1}) - [A (B + x^m-mx^m)](2)mx^{m-1}\bigg]}{(B + x^m)^4}\\ &= \frac{\bigg[(B + x^m)(A)(mx^{m-1}-m^2x^{m-1}) - [A (B + x^m-mx^m)](2)mx^{m-1}\bigg]}{(B + x^m)^3}\\ &= \frac{Amx^{m-1}\bigg[(B + x^m)(1-m) - [(B + x^m-mx^m)](2)\bigg]}{(B + x^m)^3}\\ &= \frac{Amx^{m-1}\bigg[B - mB + x^m + mx^m - 2B -2x^m+2mx^m\bigg]}{(B + x^m)^3}\\ p''(x)&=\boxed{ \frac{Amx^{m-1}\bigg[-B - mB - x^m +mx^m\bigg]}{(B + x^m)^3}}\\ \end{align*} {/eq}


We set {eq}p''(x) = 0 {/eq} and solve for x.


{eq}\displaystyle \begin{align*} p''(x) &= 0\\ \frac{Amx^{m-1}\bigg[-B - mB - x^m +mx^m\bigg]}{(B + x^m)^3} &= 0 \\ Amx^{m-1}\bigg[-B - mB - x^m +mx^m\bigg]&= 0((B + x^m)^3) \\ Amx^{m-1}\bigg[-B - mB - x^m +mx^m\bigg]&= 0\\ -B - mB - x^m +mx^m &= \frac{0}{Amx^{m-1}}\\ -B - mB - x^m +mx^m &= 0\\ -x^m +mx^m &= B + mB\\ x^m(-1 +m)&= B + mB\\ x^m&= \frac{ B + mB}{(m-1)}\\ x &=\boxed{ \sqrt[m]{\frac{ B (m+1)}{(m-1)}}}\\ \end{align*} {/eq}


Learn more about this topic:

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