N(2x+y) = x +1
Differentiate both sides, using the Chain Rule on the left side.
(1 / (2x + y)) * d(2x + y)/dx = 1
(1 / (2x + y)) * (2 + dy/dx) = 1
<span>Rearrange to isolate dy/dx.
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The answer is
Final result :
x - 2
Step by step solution :
Step 1 :
1
Simplify —
3
Equation at the end of step 1 :
2 1
((—•x)-7)+((—•x)+5)
3 3
Step 2 :
Rewriting the whole as an Equivalent Fraction :
2.1 Adding a whole to a fraction
Rewrite the whole as a fraction using 3 as the denominator :
5 5 • 3
5 = — = —————
1 3
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole
Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator
Adding fractions that have a common denominator :
2.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
x + 5 • 3 x + 15
————————— = ——————
3 3
Equation at the end of step 2 :
2 (x + 15)
((— • x) - 7) + ————————
3 3
Step 3 :
2
Simplify —
3
Equation at the end of step 3 :
2 (x + 15)
((— • x) - 7) + ————————
3 3
Step 4 :
Rewriting the whole as an Equivalent Fraction :
4.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using 3 as the denominator :
7 7 • 3
7 = — = —————
1 3
Adding fractions that have a common denominator :
4.2 Adding up the two equivalent fractions
2x - (7 • 3) 2x - 21
———————————— = ———————
3 3
Equation at the end of step 4 :
(2x - 21) (x + 15)
————————— + ————————
3 3
Step 5 :
Adding fractions which have a common denominator :
5.1 Adding fractions which have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
(2x-21) + (x+15) 3x - 6
———————————————— = ——————
3 3
Step 6 :
Pulling out like terms :
6.1 Pull out like factors :
3x - 6 = 3 • (x - 2)
Final result :
x - 2
100% Verified!
Hope This Helps! :)
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