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Chapter 5 Class 12 Continuity and Differentiability
Ex 5.1, 13
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Ex 5.2, 5
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Ex 5.3, 14
Example 29 Important
Example 30 Important You are here
Ex 5.5,6 Important
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Ex 5.6, 7 Important
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Example 38 Important
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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
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Chapter 5 Class 12 Continuity and Differentiability
Last updated at June 2, 2023 by Teachoo
Example 30 Find ππ¦/ππ₯ , if π¦^π₯+π₯^π¦+π₯^π₯=π^π. Let u = π¦π₯, v = π₯π¦ & w = π₯^π₯ Now, π + π + π = π^π Differentiating π€.π.π‘.π₯ (π (π’ + π£ + π€))/ππ₯ = (π(π^π))/ππ₯ (π(π’))/ππ₯ + (π(π£))/ππ₯ + (π(π€))/ππ₯ = 0 We will calculate derivative of u, v & w separately . Finding Derivative of π . π’ = π¦^π₯ Taking log both sides logβ‘π’=logβ‘γ (π¦^π₯)" " γ logβ‘π’=γπ₯ . logγβ‘π¦" " Differentiating both sides π€.π.π‘.π₯ (π(logβ‘π’))/ππ₯ = (π(π₯ . logβ‘π¦))/ππ₯ (π(logβ‘π’))/ππ₯ (ππ’/ππ’) = π(π₯.logβ‘π¦ )/ππ₯ 1/π’ . ππ’/ππ₯ = (π (π₯ . logβ‘π¦ ))/ππ₯ (π΄π logβ‘γ(π^π)γ=π logβ‘π) By product Rule (uv)β = uβv + vβu 1/π’ . ππ’/ππ₯ = ππ₯/ππ₯ . logβ‘π¦ + (π(logβ‘π¦))/ππ₯ . π₯ 1/π’ . ππ’/ππ₯ = 1 . logβ‘π¦ + π₯. π(logβ‘π¦ )/ππ₯ . ππ¦/ππ¦ 1/π’ . ππ’/ππ₯ = logβ‘π¦ + π₯. π(logβ‘π¦ )/ππ₯ . ππ¦/ππ₯ 1/π’ . ππ’/ππ₯ = logβ‘π¦ + π₯. 1/π¦ . ππ¦/ππ₯ 1/π’ . ππ’/ππ₯ = logβ‘π¦ + π₯/π¦ . ππ¦/ππ₯ ππ’/ππ₯ = π’ (logβ‘π¦ "+ " π₯/π¦ " " ππ¦/ππ₯) π π/π π = π^π (πππβ‘π "+ " π/π " " π π/π π) Finding derivative of v v = xy Taking log both sides logβ‘π£=logβ‘γ (π₯^π¦)" " γ logβ‘π£=γπ¦. logγβ‘π₯" " Differentiating both sides π€.π.π‘.π₯ (π(logβ‘π£))/ππ₯ = (π(π¦ . logβ‘π₯))/ππ₯ (π(logβ‘π£))/ππ₯ (ππ£/ππ₯) = π(γπ¦ logγβ‘π₯ )/ππ₯ 1/π£ (ππ£/ππ₯) = ( π(γπ¦ logγβ‘π₯ ))/ππ₯ By product Rule (uv)β = uβv + vβu 1/π£ (ππ£/ππ₯) = ( π(π¦))/ππ₯ . logβ‘π₯ + (π (logβ‘π₯))/ππ₯ . π¦ 1/π£ (ππ£/ππ₯) = ( π(π¦))/ππ₯ . logβ‘π₯ + (π (logβ‘π₯))/ππ₯ . π¦ 1/π£ (ππ£/ππ₯) = ( ππ¦)/ππ₯ . logβ‘π₯ + 1/π₯ . π¦ 1/π£ (ππ£/ππ₯) = ( ππ¦)/ππ₯ logβ‘π₯ + π¦/π₯ ππ£/ππ₯ = v (log ( ππ¦)/ππ₯ π₯+π¦/π₯) Putting values of π£ = π₯^π¦ π π/π π = π^π (π π/π π πππβ‘γπ+ π/πγ ) Calculating derivative of π π€ = π₯^π₯ Taking log both sides logβ‘π€=logβ‘γ (π₯^π₯)" " γ logβ‘π€=γπ₯. logγβ‘π₯" " Differentiating both sides π€.π.π‘.π₯ (π(logβ‘π€))/ππ₯ = (π(π₯ . logβ‘π₯))/ππ₯ (π(logβ‘π€))/ππ₯ (ππ€/ππ€) = π(π₯ logβ‘π₯ )/ππ₯ (π(logβ‘π€))/ππ€ . ππ€/ππ₯ = π(π₯ logβ‘π₯ )/ππ₯ (π΄π logβ‘γ(π^π)γ=π logβ‘π) 1/π€ . ππ€/ππ₯ = π(π₯ logβ‘π₯ )/ππ₯ logβ‘π€=γπ₯. logγβ‘π₯" " Differentiating both sides π€.π.π‘.π₯ (π(logβ‘π€))/ππ₯ = (π(π₯ . logβ‘π₯))/ππ₯ (π(logβ‘π€))/ππ₯ (ππ€/ππ€) = π(π₯ logβ‘π₯ )/ππ₯ (π(logβ‘π€))/ππ€ . ππ€/ππ₯ = π(π₯ logβ‘π₯ )/ππ₯ 1/π€ . ππ€/ππ₯ = π(π₯ logβ‘π₯ )/ππ₯ By product Rule (uv)β = uβv + vβu 1/π€ (ππ€/ππ₯) = ( π(π₯))/ππ₯ . logβ‘π₯ + (π (logβ‘π₯))/ππ₯ . π₯ 1/π€ (ππ€/ππ₯) = 1 . logβ‘π₯ + 1/π₯ . π₯ 1/π€ (ππ€/ππ₯) = (logβ‘γπ₯+1γ) ππ€/ππ₯ = π€(logβ‘γπ₯+1γ) π π/π π = π^π (πππβ‘γπ+πγ ) From (1) ππ’/ππ₯ + ππ£/ππ₯ + ππ€/ππ₯ = 0 Putting values from (2), (3) & (4) (π¦^π₯ logβ‘γπ¦+π¦^(π₯β1). π₯ ππ¦/ππ₯ γ ) + (π₯^π¦ logβ‘γπ₯.ππ¦/ππ₯+π₯^π¦.π¦/π₯ γ ) + (π₯^π₯ (logβ‘γπ₯+1γ))=0(π¦^π₯ logβ‘γπ¦+π₯^π¦. π¦/π₯+π₯^π₯ (logβ‘γπ₯+1γ)γ ) + (π¦^(π₯β1) .β‘γπ₯ ππ¦/ππ₯+π₯^π¦ logβ‘γπ₯ ππ¦/ππ₯γ γ ) = 0 (π¦^(π₯β1) .β‘γπ₯ ππ¦/ππ₯+π₯^π¦ logβ‘γπ¦ ππ¦/ππ₯γ γ ) = β (π¦^π₯ logβ‘γπ¦+π₯^π¦. π¦/π₯+π₯^π₯ (logβ‘γπ₯+1γ)γ ) (π¦^(π₯β1) .β‘γπ₯ +π₯^π¦ logβ‘γπ₯ γ γ ) ππ¦/ππ₯ = β (π¦^π₯ logβ‘γπ¦+π₯^π¦. π¦/π₯+π₯^π₯ (logβ‘γπ₯+1γ)γ ) ππ¦/ππ₯ = "β" (π¦^π₯ πππβ‘γπ¦ + π₯^π¦. π¦/π₯ + π₯^π₯ (1 + πππβ‘π₯)γ )/((γπ₯π¦γ^(π₯β1) +β‘γπ₯^π¦ πππβ‘γπ₯ γ γ)) π π/π π = "β" (π^π πππβ‘γπ + π^(π β π) π + π^π (π + πππβ‘π)γ )/((γππγ^(πβπ) +β‘γπ^π πππβ‘γπ γ γ))