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Chapter 5 Class 12 Continuity and Differentiability
Ex 5.1, 13
Ex 5.1, 16
Ex 5.1, 18 Important
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Ex 5.1, 30 Important
Ex 5.1, 34 Important
Ex 5.2, 5
Ex 5.2, 9 Important
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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 You are here
Ex 5.5, 11 Important
Ex 5.5, 16 Important
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, 7 Differentiate the functions in, γ(logβ‘γπ₯)γγ^π₯ + π₯^logβ‘π₯ Let π¦ = γ(logβ‘γπ₯)γγ^π₯+ π₯^logβ‘π₯ Let π’ = γ(logβ‘γπ₯)γγ^π₯ , π£ = π₯^logβ‘π₯ π¦ = π’+π£ Differentiating both sides π€.π.π‘.π₯. ππ¦/ππ₯ = (π (π’ + π£))/ππ₯ ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Calculating π π/π π π’ = γ(logβ‘γπ₯)γγ^π₯ Taking log both sides logβ‘π’ = log γ(logβ‘γπ₯)γγ^π₯ logβ‘π’ = π₯ . log (logβ‘γπ₯)γ Differentiating both sides π€.π.π‘.π₯. π(logβ‘π’ )/ππ’ . ππ’/ππ₯ = (π(π₯ . log (logβ‘γπ₯)γ ) )/ππ₯ 1/π’ (ππ’/ππ₯)" =" π(π₯)/ππ₯ . log (log" " π₯) + π(log (log" " π₯))/ππ₯ Γ π₯ Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ 1/π’ (ππ’/ππ₯)" =" 1 . log (log" " π₯) + (1/(log" " π₯) .π(log" " π₯)/ππ₯) Γ π₯ 1/π’ (ππ’/ππ₯)" =" log (log" " π₯) + (1/(log" " π₯) . 1/π₯) Γ π₯ 1/π’ (ππ’/ππ₯)" =" log (log" " π₯) + 1/logβ‘π₯ Γ π₯/π₯ 1/π’ (ππ’/ππ₯)" =" log (log" " π₯) + 1/logβ‘π₯ ππ’/ππ₯ " =" π’ (log (log" " π₯)" + " 1/logβ‘π₯ ) ππ’/ππ₯ = (logβ‘π₯ )^π₯ (log (log" " π₯)" + " 1/logβ‘π₯ ) ππ’/ππ₯ = (logβ‘π₯ )^π₯ ((logβ‘γπ₯ . γlog γβ‘(logβ‘π₯ ) +γ 1)/logβ‘π₯ ) ππ’/ππ₯ = (logβ‘π₯ )^π₯/logβ‘π₯ (logβ‘γπ₯ . γlog γβ‘(logβ‘π₯ )+γ 1) ππ’/ππ₯ = (logβ‘π₯ )^(π₯ β1) (logβ‘γπ₯ . γlog γβ‘(logβ‘π₯ )+γ 1) Calculating π π/π π π£ = π₯^logβ‘π₯ Taking log both sides . logβ‘π£=logβ‘γ (π₯^logβ‘π₯ )γ logβ‘π£ = log π₯ . logβ‘π₯ logβ‘π£ = (logβ‘π₯ )^2 (As πππβ‘(π^π )=π . πππβ‘π) Differentiating both sides π€.π.π‘.π₯. π(logβ‘π£ )/ππ₯ = (π(logβ‘π₯ )^2)/ππ₯ π(logβ‘π£ )/ππ₯ . ππ£/ππ£ = (π(logβ‘π₯ )^2)/ππ₯ π(logβ‘π£ )/ππ£ . ππ£/ππ₯ = (π(logβ‘π₯ )^2)/ππ₯ 1/π£ . ππ£/ππ₯ = (π(logβ‘π₯ )^2)/ππ₯ 1/π£ . ππ£/ππ₯ = 2 logβ‘π₯ . π(logβ‘π₯ )/ππ₯ 1/π£ . ππ£/ππ₯ = 2 logβ‘π₯ . 1/π₯ ππ£/ππ₯ = π£ ((2 logβ‘π₯)/π₯) ππ£/ππ₯ = π₯^logβ‘π₯ ((2 logβ‘π₯)/π₯) ππ£/ππ₯ = π₯^logβ‘π₯ /π₯ (2 logβ‘π₯ ) ππ£/ππ₯ = π₯^(logβ‘π₯ β 1) . 2 logβ‘π₯ ππ£/ππ₯ = γ2π₯γ^(logβ‘π₯ β 1) . logβ‘π₯ Hence ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Putting value of ππ’/ππ₯ & ππ£/ππ₯ π π/π π = (π₯π¨π β‘π )^(π βπ) (π+π₯π¨π β‘π.π₯π¨π β‘(π₯π¨π β‘π ) ) + γππγ^(πππ π βπ). πππβ‘π