There are a number of ways you can write the piecewise function. Here is one of them:

$p(t)=\left\{\begin{array}{rcl}7t&\mbox{for}&t\leq40\\10.5t-140&\mbox{for}&t>40\end{array}\right.$

It could also be something like
p(t) = 7t + 3.50×max(0, t-40)
where the max(x,y) function returns the larger of x or y.

5 0

2 years ago

Differentiate y = x5√x. SOLUTION Since both the base and the exponent are variable, we use logarithmic differentiation: ln y = l

AnnZ [28]

Answer:

$y'=x^{5\sqrt{x} } (\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)$

Step-by-step explanation:

Given: $y=x^{5\sqrt{x} }$

Take ln of both sides

$lny=lnx^{5\sqrt{x} } = 5\sqrt{x}lnx\\\\lny=5\sqrt{x}lnx$

Differentiate with respect to x

$\frac{d}{dx}(lny)=\frac{d}{dx}(5\sqrt{x}lnx)\\\\\frac{d}{dy}(lny)*\frac{dy}{dx}=5\sqrt{x}\frac{d}{dx}lnx+\frac{d}{dx}5\sqrt{x}*lnx)\\\\\frac{1}{y}*\frac{dy}{dx}=(5\sqrt{x}*\frac{1}{x})+(5*\frac{1}{2}x^{-\frac{1}{2}}lnx)\\ \\\frac{1}{y}*y^{'}=\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx \\\\y'=y(\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)\\\\y'=x^{5\sqrt{x} } (\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)$

8 0

2 years ago