Find the partial derivative with respect to t of d = (Vy · 100)/(Vm + Vm20 · (t − 20) · b) − 100, treating Vy, Vm, Vm20 and b as constants.

Step-by-step solution

1. 1

   Drop the additive constant and name the denominator. The trailing $-100$ has zero derivative, so it plays no part. Setting

   $D=V_m+V_{m20}(t-20)b,$

   the function becomes

   $d=100V_y\,D^{-1}-100.$

   Writing the quotient as a negative power turns a quotient rule into a one-line chain rule.

2. 2

   Differentiate $D$ with respect to $t$. Only the factor $(t-20)$ depends on $t$, and its derivative is $1$:

   $\frac{\partial D}{\partial t}=V_{m20}b.$

   Note the $-20$ is a shift, not a scale, so it does not appear in the derivative.

3. 3

   Apply the chain rule to $D^{-1}$.

   $\frac{\partial}{\partial t}\bigl(D^{-1}\bigr)=-D^{-2}\frac{\partial D}{\partial t}=-\frac{V_{m20}b}{D^{2}}.$

4. 4

   Multiply by the constant factor.

   $\frac{\partial d}{\partial t}=100V_y\left(-\frac{V_{m20}b}{D^{2}}\right)=-\frac{100\,V_y\,V_{m20}\,b}{\bigl(V_m+V_{m20}(t-20)b\bigr)^{2}}.$

5. 5

   Interpret the sign. The square in the denominator is positive, so the sign of $\partial d/\partial t$ is the opposite of the sign of $V_y V_{m20} b$ and never flips as $t$ changes. If the expansion coefficient $b$ is zero the derivative vanishes identically, which is right: with $b=0$ the variable $t$ disappears from the formula altogether.