Derive the function z/(e^(-1x)) with respect to x

\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\frac{z}{e^{\left(-1\right)\cdot x}}\right)

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Answer

$0=\frac{ze^{-x}}{e^{-2x}}$

Step by step solution

Problem

$\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\frac{z}{e^{\left(-1\right)\cdot x}}\right)$
1

The derivative of the constant function is equal to zero

$0=\frac{d}{dx}\left(\frac{z}{e^{-x}}\right)$
2

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$0=\frac{e^{-x}\cdot\frac{d}{dx}\left(z\right)-z\frac{d}{dx}\left(e^{-x}\right)}{\left(e^{-x}\right)^2}$
3

The derivative of the constant function is equal to zero

$0=\frac{0e^{-x}-z\frac{d}{dx}\left(e^{-x}\right)}{\left(e^{-x}\right)^2}$
4

Any expression multiplied by $0$ is equal to $0$

$0=\frac{0-z\frac{d}{dx}\left(e^{-x}\right)}{\left(e^{-x}\right)^2}$
5

Applying the derivative of the exponential function

$0=\frac{0-1\cdot 1z\frac{d}{dx}\left(-x\right)e^{-x}}{\left(e^{-x}\right)^2}$
6

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$0=\frac{0-1\cdot 1\left(-1\right)ze^{-x}\cdot\frac{d}{dx}\left(x\right)}{\left(e^{-x}\right)^2}$
7

The derivative of the linear function is equal to $1$

$0=\frac{0-1\cdot 1\cdot 1\left(-1\right)ze^{-x}}{\left(e^{-x}\right)^2}$
8

Multiply $-1$ times $-1$

$0=\frac{1ze^{-x}+0}{\left(e^{-x}\right)^2}$
9

$x+0=x$, where $x$ is any expression

$0=\frac{ze^{-x}}{\left(e^{-x}\right)^2}$
10

Applying the power of a power property

$0=\frac{ze^{-x}}{e^{-2x}}$
11

Rewriting the fraction

$0=z\frac{1}{e^{-2x}}e^{-x}$
12

Multiplying the fraction and term

$0=\frac{ze^{-x}}{e^{-2x}}$

Answer

$0=\frac{ze^{-x}}{e^{-2x}}$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.24 seconds

Views:

103