Given e^(x+y) – x=0 the dx/dy is equal to?

Question

a) 11/e^x + y
b) 1/e^(x+y)
c) 1- e^(x+y)/e^(x+y)
d) 1- e^(x+y)/e^x

Answer 1

e^(x+y) - x = 0



Take the derivative with respect to x

e^(x+y) ( 1+ dy/dx) - 1 = 0

e^(x+y) + dy/dx e^(x+y) = 1

dy/dx e^(x+y) = 1- e^(x+y)

dy/dx = ( 1- e^(x+y) ) / e^(x+y)



The answer is (c) but you need to insert parenthesis to avoid confusion.

Answer 2

e^(x+y) – x=0

=> d/dx(e^(x+y) – x) = d/dx(0)

term by term

=> d/dx(e^(x+y)) – d/dx(x) = d/dx(0)

given the chain rule d/dx(e^u) = e^u * d/dx(u)

=> e^(x+y) * d/dx(x + y) – d/dx(x) = d/dx(0)

term by term

=> e^(x+y) * (d/dx(x) + d/dx(y)) – d/dx(x) = d/dx(0)

using the power rule and d/dx(y) = dy/dx

=> e^(x+y) * (1 + dy/dx) – 1 = 0

=> e^(x+y) * (1 + dy/dx) = 1

=> e^(x+y) + (dy/dx)e^(x+y) = 1

=> (dy/dx)e^(x+y) = 1 – e^(x+y)

=> dy/dx = (1 – e^(x+y))/e^(x+y) <== answer as a single term

=> dy/dx = 1/e^(x+y) – e^(x+y))/e^(x+y)

=> dy/dx = 1/e^(x+y) – 1

=> dy/dx = e^(-x–y) – 1 <== answer simplified.



I it possible that Answer C is correct if you meant to write (1 – e^(x+y))/e^(x+y)

Answer 3

e^(x+y) - x = 0

differentiate y w.r.t x

e^(x+y).[1+dy/dx] - 1 =0

e^(x+y) + e^(x+y) dy/dx = 1

e^(x+y) dy/dx = 1 - e^(x+y)

dy/dx = [1 - e^(x+y)]/e^(x+y)

further the answer can be

dy/dx = (1-x)/x [as e^(x+y) = x]

Answer 4

e^(x+y) – x=0

e^(x + y) (1 + dy/dx) - 1 = 0

e^(x + y) + e^(x + y)dy/dx - 1 = 0

e^(x + y)dy/dx = 1 - e^(x + y)



..............1 - e^(x + y)

dy/dx = ------------------------ Answer//

.................e^(x + y)



Answer is c) 1- e^(x+y)/e^(x+y)

Answer 5

Rewrite e^(x + y) as (e^y)(e^x):



(e^y)(e^x) - x = 0



To implicitly differentiate this, one must uses the product rule and the chain rule on the first term and one uses the power rule on the second term:



(e^y)(dy/dx)(e^x) + (e^y)(e^x) - 1 = 0





Move all the terms that do NOT contain dy/dx to the right:



(e^y)(dy/dx)(e^x) = 1 - (e^y)(e^x)



Put them back to x + y form:



(dy/dx){e^(x + y)} = 1 - e^(x + y)



Divide by e^(x + y):



dy/dx = {1 - e^(x + y)}/e^(x + y)

Answer 6

You'll need the chain rule and implicit differentiation. Start with taking d/dx of both sides (using ' for derivative w.r.t. x):



e^(x+y) (x+y)' - 1 = 0

e(x+y) (1 + y') = 1 .... added 1 to both sides and simplified (x+y)'



Solve for y' and you get:



y' = (1 / e^(x + y)) - 1 = [1 - e^(x+y)] / e^(x+y)