Exponential & Logarithms (Edexcel International A Level (IAL) Maths: Pure 3): Flashcards

Exam code: YMA01

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If $y = e^{k x}$ then $\frac{d y}{d x} =______$ , and if $y = e^{- k x}$ then $\frac{d y}{d x} =______$ .
If $y = e^{k x}$ then $\frac{d y}{d x} = k e^{k x}$ , and if $y = e^{- k x}$ then $\frac{d y}{d x} = - k e^{- k x}$ .
The constant from the power comes down as a multiplier, and $y = e^{x}$ is simply the case $k = 1$ .
What stays the same when you differentiate $e^{k x}$ ?
The exponential part itself: $e^{k x}$ reappears in the derivative unchanged.
Only a constant multiplier is added in front, which is what makes exponential derivatives unusually simple.
What is the derivative of $y = e^{- 3 x}$ ?
$\frac{d y}{d x} = - 3 e^{- 3 x}$ .
The minus sign comes down with the $3$ , so a decay curve has a negative gradient everywhere along it.
True or False?
The gradient of $y = e^{x}$ is never zero.
True.
The gradient equals $e^{x}$ , which is positive for every value of $x$ .
So the curve is always increasing and has no stationary points at all.
How do you find the gradient of $y = e^{2 x}$ at $x = 0$ ?
Differentiate to get $\frac{d y}{d x} = 2 e^{2 x}$ , then substitute $x = 0$ .
Since $e^{0} = 1$ , the gradient there is $2$ .

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