If y = 2x 32x 5–x 7–x then (\frac{dy}{dx}) = | MATH - DIFFERENTIATION BY SUBSTITUTION, HIGHER ORDER DERIVATIVES | SolveeIt

If y = 2^x 3^2x 5^–x 7^–x then $\frac{dy}{dx}$ =

$\frac{2^{x}.3^{2x}}{5^{x}7^{x}}\log\frac{18}{35}$

$\frac{2^{x}.3^{2x}}{5^{x}7^{x}}\log\frac{35}{18}$

$\frac{2^{x}.3^{2x}}{5^{x}7^{x}}\log\frac{14}{45}$

$\frac{2^{x}.3^{2x}}{5^{x}7^{x}}\log\frac{45}{14}$

Answer

$\frac{2^{x}.3^{2x}}{5^{x}7^{x}}\log\frac{18}{35}$

Explanation

y = $\frac{(18)^{x}}{(35)^{x}}$ ̃ y = $\left( \frac{18}{35} \right)^{x}$

$\frac{dy}{dx}$ = $\left( \frac{18}{35} \right)^{x}\log\frac{18}{35}$

Question: If y = 2<sup>x</sup> 3<sup>2x</sup> 5<sup>–x</sup> 7<sup>–x</sup> then \(\frac{dy}{dx}\) =...