Check sibling questions

if e x + e y =e (x+y) , then dy/dx is :

(a) e (y-x) Β  Β  Β (b) e (x+y)
(c) -e (y-x) Β  Β  (d) 2 e (x-y)



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Question 12 if 𝑒^π‘₯+ 𝑒^𝑦=𝑒^(π‘₯+𝑦), π‘‘β„Žπ‘’π‘› 𝑑𝑦/𝑑π‘₯ is : (a) 𝑒^(π‘¦βˆ’π‘₯) (b) 𝑒^(π‘₯+𝑦) (c) γ€–βˆ’π‘’γ€—^(π‘¦βˆ’π‘₯) (d) 2 𝑒^(π‘₯βˆ’π‘¦) Given 𝑒^π‘₯+𝑒^𝑦=𝑒^(π‘₯ + 𝑦) 𝑒^π‘₯+𝑒^𝑦=𝑒^π‘₯.𝑒^𝑦 Dividing by 𝒆^𝒙.𝒆^π’š both sides 𝒆^𝒙/(𝒆^𝒙 𝒆^π’š )+𝒆^π’š/(𝒆^𝒙 𝒆^π’š )=(𝒆^𝒙 𝒆^π’š)/(𝒆^𝒙 𝒆^π’š ) 1/𝑒^𝑦 +1/𝑒^π‘₯ =1 𝒆^(βˆ’π’š)+𝒆^(βˆ’π’™)=𝟏 Differentiating both sides w.r.t x 𝒅(𝒆^(βˆ’π’š) )/𝒅𝒙+𝒅(𝒆^(βˆ’π’™) )/𝒅𝒙=𝒅(𝟏)/𝒅𝒙 𝒆^(βˆ’π’š).(𝒅(βˆ’π’š))/π’…π’™βˆ’π’†^(βˆ’π’™)=𝟎 βˆ’π‘’^(βˆ’π‘¦).𝑑𝑦/𝑑π‘₯βˆ’π‘’^(βˆ’π‘₯)=0 βˆ’π‘’^(βˆ’π‘¦).𝑑𝑦/𝑑π‘₯=𝑒^(βˆ’π‘₯) π’…π’š/𝒅𝒙=𝒆^(βˆ’π’™)/(βˆ’π’†^(βˆ’π’š) ) 𝑑𝑦/𝑑π‘₯=βˆ’π‘’^(βˆ’π‘₯βˆ’(βˆ’π‘¦)) 𝑑𝑦/𝑑π‘₯=βˆ’π‘’^(βˆ’π‘₯ + 𝑦) π’…π’š/𝒅𝒙=βˆ’π’†^(π’š βˆ’ 𝒙) So, the correct answer is (c)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.