#### To determine

**To find:** The value of dvd[S].

#### Answer

The derivative of the function v=0.14[S]0.015+[S] is dvd[S]=0.0021[0.015+[S]]2.

The terms dvd[S] means that the rate of change of the rate of an enzymatic reaction with respect to the concentration of a substrate *S*.

#### Explanation

**Given:**

The equation for the enzyme chymotrypsin is v=0.14[S]0.015+[S]

**Derivative rule:**

(1) Quotient Rule: If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]−f1(x)ddx[f2(x)][f2x]2

(2) Power rule: ddx(xn)=nxn−1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

(4) Constant multiple rule: ddx(c⋅f)=c⋅ddx(f)

**Calculation:**

The derivative of the function v=0.14[S]0.015+[S] is dvd[S] which is obtained as follows,

dvd[S]=dd[S](0.14[S]0.015+[S])

Apply the Quotient rule (1) and simplify the terms,

dvd[S]=(0.015+[S])⋅dd[S](0.14[S])−(0.14[S])⋅dd[S](0.015+[S])[0.015+[S]]2

Apply the sum rule (3) and constant multiple rule (4),

dvd[S]=[(0.015+[S])[(0.14)⋅dd[S]([S])]]−[(0.14[S])[dd[S](0.015)+dd[S]([S])]][0.015+[S]]2

Apply the power rule (2) and simplify the terms,

dvd[S]=[(0.015+[S])[(0.14)(1)]]−[(0.14[S])[0+1]][0.015+[S]]2=(0.015+[S])(0.14)−0.14[S][0.015+[S]]2=0.0021+0.14[S]−0.14[S][0.015+[S]]2=0.0021[0.015+[S]]2

Therefore, the derivative of the function v=0.14[S]0.015+[S] is dvd[S]=0.0021[0.015+[S]]2.

**To describe:** The meaning of the term dvd[S].

**Explanation:**

Given that, *v* is the rate of an enzymatic reaction.

Therefore, the terms dvd[S] means that the rate of change of the rate of an enzymatic reaction with respect to the concentration of a substrate [S].