$\displaystyle h(x)=\int\limits_{1}^{\sqrt{x}}~\cfrac{z^2}{z^4+1}dz~\hspace{10em}\cfrac{dh}{du}\cdot \stackrel{chain~rule}{\cfrac{du}{dx}\implies \cfrac{dh}{dx}} \\\\[-0.35em] ~\dotfill\\\\ u=\sqrt{x}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}} \\\\[-0.35em] ~\dotfill$

$\cfrac{dh}{dx}\implies \displaystyle \cfrac{d}{du}\left[ \int\limits_{1}^{u}~\cfrac{z^2}{z^4+1}dz \right]\cdot \cfrac{1}{2\sqrt{x}}\implies \left[ \cfrac{u^2}{u^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}} \\\\\\ \stackrel{substituting~back}{\left[ \cfrac{(\sqrt{x})^2}{(\sqrt{x})^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}}}\implies \cfrac{x}{x^2+1}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{\sqrt{x}}{2x^2+2}$

7 0

1 year ago

Consider this function for cell duplication where the cells duplicate every minute.

astraxan [27]

Answer:

A) The 75 is the initial number of cells, and the 2 indicates that the number of cells doubles every minute

Step-by-step explanation:

When x=0 (no minutes have elapsed), the value of the function is ...

f(0) = 75(2^0) = 75(1) = 75 . . . . the initial number of cells

As x increases by 1 (minute), the number of cells is multiplied by 2, so the 2 is the multiplier each minute. It indicates the number doubles. ("double" = "multiply by 2")

x is not defined, but for the function to make any sense, it must represent elapsed minutes.

3 0

2 years ago