Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Chapter 5 Class 12 Continuity and Differentiability
Ex 5.1, 13
Ex 5.1, 16
Ex 5.1, 18 Important
Ex 5.1, 28 Important
Ex 5.1, 30 Important
Ex 5.1, 34 Important
Ex 5.2, 5
Ex 5.2, 9 Important
Ex 5.2, 10 Important
Ex 5.3, 10 Important
Ex 5.3, 14
Example 29 Important
Example 30 Important
Ex 5.5,6 Important
Ex 5.5, 7 Important
Ex 5.5, 11 Important
Ex 5.5, 16 Important You are here
Ex 5.6, 7 Important
Ex 5.6, 11 Important
Example 38 Important
Ex 5.7, 14 Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Example 39 (i)
Example 40 (i)
Example 42 Important
Misc 6 Important
Misc 15 Important
Misc 16 Important
Misc 22 Important
Chapter 5 Class 12 Continuity and Differentiability
Last updated at May 29, 2023 by Teachoo
Ex 5.5, 16 Find the derivative of the function given by f (π₯) = (1 + π₯) (1 + π₯^2) (1 + π₯^4) (1 + π₯8) and hence find f β²(1) .Given π(π₯)=(1+π₯)(1+π₯^2 )(1+π₯^4 )(1+π₯^8 )" " Let π¦=(1+π₯)(1+π₯^2 )(1+π₯^4 )(1+π₯^8 ) Taking log both sides log π¦ = log (1+π₯)(1+π₯^2 )(1+π₯^4 )(1+π₯^8 ) log π¦ = log (1+π₯)+logβ‘(1+π₯^2 )+logβ‘(1+π₯^4 ) γ+ logγβ‘γ (1+π₯^8 )γ Differentiating both sides π€.π.π‘.π₯. π(logβ‘π¦ )/ππ₯ = π(log (1 + π₯) + logβ‘(1 + π₯^2 ) + logβ‘(1 + π₯^4 )+ logβ‘γ (1 + π₯^8 )γ )/ππ₯ π(logβ‘π¦ )/ππ₯ = π(log (1 + π₯))/ππ₯ + π(logβ‘(1 + π₯^2 ) )/ππ₯ + π(logβ‘(1 + π₯^4 ) )/ππ₯ + π(logβ‘γ (1 + π₯^8 )γ )/ππ₯ π(logβ‘π¦ )/ππ¦ . ππ¦/ππ₯ = 1/(1 + π₯) . π(1 + π₯)/ππ₯ + 1/((1 + π₯^2 ) ) . π(1 + π₯^2 )/ππ₯ + 1/((1 + π₯^4 ) ) . π(1 + π₯^4 )/ππ₯ + 1/((1 + π₯^8 ) ) . π(1 + π₯^8 )/ππ₯ 1/π¦ . ππ¦/ππ₯ = 1/(1 + π₯) . (0+1) + 1/((1 + π₯^2 ) ) . (0+2π₯) + 1/((1 + π₯^4 ) ) . (0+4π₯^3 ) + 1/((1 + π₯^8 ) ) . (0+8π₯^7 ) 1/π¦ . ππ¦/ππ₯ = 1/(1 + π₯) + 2π₯/(1 + π₯^2 ) + (4π₯^3)/(1 + π₯^4 ) + (8π₯^7)/(1 + π₯^8 ) ππ¦/ππ₯ = π¦ (1/(1 + π₯) " + " 2π₯/(1 + π₯^2 ) " + " (4π₯^3)/(1 + π₯^4 ) " + " (8π₯^7)/(1 + π₯^8 )) ππ¦/ππ₯ = (1+π₯)(1+π₯^2 )(1+π₯^4 )(1+π₯^8 ) (1/(1 + π₯) " + " 2π₯/(1 + π₯^2 ) " + " (4π₯^3)/(1 + π₯^4 ) " +" (8π₯^7)/(1 + π₯^8 )) Hence, πβ²(π) = (π+π)(π+π^π )(π+π^π )(π+π^π ) (π/(π + π) " + " ππ/(π + π^π ) " + " (ππ^π)/(π + π^π ) " + " (ππ^π)/(π + π^π )) We need to find πβ²(1) Putting π₯=1 πβ²(1) = (1+1)(1+(1)^2 )(1+(1)^4 )(1+γ(1)γ^8 ) (1/(1 +1) " + " 2(1)/(1+(1)^2 ) " + " (4(1)^3)/(1 + (1)^4 ) " + " (8(1)^7)/(1 + (1)^8 )) = 2(1+1)(1+1)(1+1) (1/(1 + 1) " + " 2/(1 + 1) " + " 4/(1 + 1) " + " 8/(1 + 1)) = 2(2)(2)(2) (1/2 " + " 2/2 " + " 4/2 " + " 8/2) ππ¦/ππ₯ = (1+π₯)(1+π₯^2 )(1+π₯^4 )(1+π₯^8 ) (1/(1 + π₯) " + " 2π₯/(1 + π₯^2 ) " + " (4π₯^3)/(1 + π₯^4 ) " +" (8π₯^7)/(1 + π₯^8 )) Hence, πβ²(π) = (π+π)(π+π^π )(π+π^π )(π+π^π ) (π/(π + π) " + " ππ/(π + π^π ) " + " (ππ^π)/(π + π^π ) " + " (ππ^π)/(π + π^π )) We need to find πβ²(1) Putting π₯=1 πβ²(1) = (1+1)(1+(1)^2 )(1+(1)^4 )(1+γ(1)γ^8 ) (1/(1 +1) " + " 2(1)/(1+(1)^2 ) " + " (4(1)^3)/(1 + (1)^4 ) " + " (8(1)^7)/(1 + (1)^8 )) = 2(1+1)(1+1)(1+1) (1/(1 + 1) " + " 2/(1 + 1) " + " 4/(1 + 1) " + " 8/(1 + 1)) = 2(2)(2)(2) (1/2 " + " 2/2 " + " 4/2 " + " 8/2)