1 Answer

Awiebe · Answer 1 · 2021-12-11T18:08:55+0000

Let y be the expression you want to differentiate:

$y=\log_7(5-3x) \implies 7^y=5-3x$

Now,

$7^y=e^(\ln(7^y))=e^(y\ln(7))$

Use the chain rule to differentiate both sides with respect to x :

$\ln(7)e^(y\ln(7))(\mathrm dy)/(\mathrm dx)=-3$

Solve for dy/dx :

$\ln(7)7^y(\mathrm dy)/(\mathrm dx)=-3$

$(\mathrm dy)/(\mathrm dx)=-\frac3{\ln(7)7^y}$

$(\mathrm dy)/(\mathrm dx)=\boxed{-\frac3{\ln(7)(5-3x)}}$

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