Question

y=2^x 3^2x 5^-x. 7^-x then dy\ dx=​

Answer 1

SOLUTION

GIVEN

$\sf{}y = {2}^{x} . {3}^{2x} . {5}^{ - x} . {7}^{ - x}$

TO DETERMINE

$\displaystyle \sf{} \frac{dy}{dx}$

FORMULA TO BE IMPLEMENTED

$\sf{} 1. \: \: {a}^{m} \times {b}^{m} = {(ab)}^{m}$

$\displaystyle \sf{}2. \: \: \frac{d}{dx} ( {a}^{x} ) = {a}^{x} \log a$

EVALUATION

Here it is given that

$\sf{}y = {2}^{x} . {3}^{2x} . {5}^{ - x} . {7}^{ - x}$

$\implies \sf{}y = {2}^{x} . {( {3}^{2} )}^{x} . { ({5}^{ - 1}) }^{ - x} . {( {7}^{ - 1}) }^{ x}$

$\displaystyle \implies \sf{}y = {2}^{x} . {9}^{x} . { \bigg( \frac{1}{5} \bigg) }^{ x} . { \bigg( \frac{1}{7} \bigg) }^{ x}$

$\displaystyle \implies \sf{}y = {(2 \times 9 )}^{x} . { \bigg( \frac{1}{5} \times \frac{1}{7} \bigg) }^{ x}$

$\displaystyle \implies \sf{}y = {(18)}^{x} . { \bigg( \frac{1}{35} \bigg) }^{ x}$

$\displaystyle \implies \sf{}y = { \bigg( \frac{18}{35} \bigg) }^{ x}$

Differentiating both sides with respect to x we get

$\displaystyle \sf{} \frac{dy}{dx} = { \bigg( \frac{18}{35} \bigg) }^{ x} \log \bigg( \frac{18}{35} \bigg)$

FINAL ANSWER

$\boxed{ \displaystyle \sf{} \: \: \frac{dy}{dx} = { \bigg( \frac{18}{35} \bigg) }^{ x} \log \bigg( \frac{18}{35} \bigg) \: \: }$

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