Given the function [tex]$y=\frac{10^x}{x+2}$[/tex] find [tex]$\frac{dy}{dx}$[/tex].
Answer (1)
Apply the quotient rule: d x d ( v u ) = v 2 v d x d u − u d x d v , where u = 1 0 x and v = x+2} .
Find the derivative of u : d x d u = 1 0 x ln ( 10 ) .
Find the derivative of v : d x d v = 1 .
Substitute and simplify: d x d y = ( x + 2 ) 2 1 0 x (( x + 2 ) l n ( 10 ) − 1 ) .
d x d y = ( x + 2 ) 2 1 0 x (( x + 2 ) ln ( 10 ) − 1 )
Explanation
Problem Analysis We are given the function y = x + 2 1 0 x and asked to find its derivative, d x d y . This requires us to use the quotient rule.
Quotient Rule The quotient rule states that if y = v u , then d x d y = v 2 v d x d u − u d x d v . In our case, u = 1 0 x and v = x + 2 .
Derivative of u First, we find the derivative of u with respect to x . We know that d x d ( a x ) = a x ln ( a ) , so d x d u = d x d ( 1 0 x ) = 1 0 x ln ( 10 ) .
Derivative of v Next, we find the derivative of v with respect to x . We have d x d v = d x d ( x + 2 ) = 1 .
Applying the Quotient Rule Now, we substitute u , v , d x d u , and d x d v into the quotient rule formula: d x d y = ( x + 2 ) 2 ( x + 2 ) ( 1 0 x ln ( 10 )) − ( 1 0 x ) ( 1 )
Simplifying the Expression Finally, we simplify the expression: d x d y = ( x + 2 ) 2 1 0 x (( x + 2 ) ln ( 10 ) − 1 )
Final Answer Therefore, the derivative of the given function is: d x d y = ( x + 2 ) 2 1 0 x (( x + 2 ) ln ( 10 ) − 1 )
Examples
Consider a scenario where you are modeling the growth of a bacterial population. The function y = x + 2 1 0 x could represent the population size at time x , where the numerator 1 0 x represents exponential growth and the denominator x + 2 represents a factor that slows down the growth due to resource limitations or other environmental factors. Finding the derivative d x d y would then give you the rate of change of the population at any given time x , which is crucial for understanding and predicting the population dynamics. This type of analysis is used in various fields such as microbiology, ecology, and epidemiology.
