What is the first differential of $y = 10^{3 x - 4}$ $y=10^(3x-4)$ ?

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What is the first differential of $y = 10^{3 x - 4}$ $y=10^(3x-4)$ ?

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Answer 1 · 2017-04-30T04:17:05

$\frac{d y}{d x} = 3 ln 10 \cdot 10^{3 x - 4}$ $dy/dx= 3ln10*10^(3x-4)$

Explanation:

$y = 10^{3 x - 4}$ $y=10^(3x-4)$

$ln y = (3 x - 4) \cdot ln 10$ $lny = (3x-4)*ln10$

$\frac{1}{y} \frac{d y}{d x} = ln 10 \cdot \frac{d}{d x} (3 x - 4)$ $1/y dy/dx = ln10*d/dx(3x-4)$ [Implicit differentiation and standard differential]

$\frac{1}{y} \frac{d y}{d x} = ln 10 \cdot (3 - 0)$ $1/y dy/dx= ln10*(3-0)$ [Power rule]

$\frac{d y}{d x} = 3 ln 10 \cdot y$ $dy/dx = 3ln10*y$

$= 3 ln 10 \cdot 10^{3 x - 4}$ $= 3ln10*10^(3x-4)$

Answer 2 · 2017-04-30T15:27:50

A few additional observations

Explanation:

Notice that $10^{3 x - 4} = \frac{10^{3 x}}{10^{4}}$ $10^(3x-4) = 10^(3x)/10^4$
So that $10^{3 x - 4} = \frac{1000^{x}}{10000}$ $10^(3x-4) = 1000^(x)/10000$
Therefore if $y = 10^{3 x - 4}$ $y = 10^(3x-4)$ then $y = \frac{1000^{x}}{10000}$ $y = 1000^(x)/10000$

This gives us $\frac{d y}{d x} = (\frac{1}{10000}) 1000^{x} \cdot ln (1000)$ $dy/dx = (1/10000)1000^x*ln(1000)$ .
If we wish to simplify ln(1000), we have
$\frac{d y}{d x} = (\frac{3 ln 10}{10000}) 1000^{x}$ $dy/dx = ((3ln10)/10000)1000^x$ .

This answer is equal to the one given in the original response.

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What is the first differential of $y = 10^{3 x - 4}$ $y=10^(3x-4)$ ?

2 Answers

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