
Differentiation
QuizΒ by Rajashree Nambiar
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π/ππ₯ [log {π π₯( π₯β2 / π₯+2 ) 3/4 }] is equal to
Derivative of 2βcot ( π₯ 2) with respect to π₯ is
Derivative of βπ βπ₯ with respect to π₯ is
If π₯ = 4π‘, π¦ = 4/π‘ , ππ¦ / ππ₯ is equal to
If π₯ = πΌ(cos ΞΈ + ΞΈ sin ΞΈ) and π¦ = πΌ(sin ΞΈ + ΞΈ cos ΞΈ), then ππ¦/ππ₯ is equal to
Derivative of βπ‘ππβπ₯Β with respect to π₯ is
If π₯ = πΌ π ππ3 ΞΈ and π¦ = πΌ π‘ππ3 ΞΈ then the value of ππ¦/ππ₯ at ΞΈ = Ο /3 is
Derivative of log[log(log π₯ 5)] with respect to π₯ is
If π₯ = π‘ + 1 /π‘ and π¦ = π‘ β 1/ π‘ , then ππ¦/ππ₯ is equal to
If π₯ = 1+log π‘ /π‘ 2 and π¦ = 3+2 log π‘ /π‘ 2 , then ππ¦/ππ₯ is equal
The value of ππ¦/ππ₯ [(π‘ππ2 2π₯βπ‘ππ2 π₯ / 1βπ‘ππ2 2π₯ π‘ππ2 π₯) cot 3π₯] is
If π¦ = ππππΒ π₯, where ππππΒ means log log logβ¦β¦β¦(repeated n times), then π₯ πππ π₯ πππ2Β π₯ πππ3π₯ β¦ β¦ β¦ = ππππβ1Β π₯ πππΒ ππ₯ ππ¦/ππ₯ is equal to
If π(π₯) = β1 + πππ 2(π₯2) , then the value of πβ² (βπ/2) is
If π(π₯) = (πππcot π₯ tan π₯)(πππcot π₯ cot π₯)β1 + π‘ππβ1Β 4π₯ / 4βπ₯2, then πβ²(π₯) is equal to
If π(π₯) = ππππ₯(ππππ₯ π₯) then πβ²(π₯) at π₯ = π is equal to
Let π(π₯) = ππ₯, π(π₯) = π ππβ1π₯ and β(π₯) = π[π(π₯)], then ββ²(π₯) / β(π₯)is equal to
Let π(π₯) = sin π₯ , π(π₯) = π₯2 and β(π₯) = πππππ₯. If πΉ(π₯) = (βππππ)(π₯), then πΉβ²β²(π₯) is equal to
If π₯π¦ + π¦2 = tan π₯ + π¦, π‘βππ ππ¦/ππ₯ is equal to
If π ππ2π₯ + πππ 2π¦ = 1, then ππ¦/ππ₯ is equal to
If π₯β1 + π¦ + π¦β1 + π₯ = 0 for β1 < π₯ < 1, then ππ¦/ππ₯ is
Q.If (π₯ β π) 2 + (π¦ β π) 2 = π 2 , for some c > 0, then [1+( ππ¦ /ππ₯) 2 ] 3/2 / π2π¦ / ππ₯2 is
If cos π¦ = π₯ cos (π + π¦), with cos πΌ β 1 ,then ππ¦/ππ₯ is equal to
If x sin(π + π¦) + sin πΌ cos (π + π¦) = 0, then ππ¦/ππ₯ is equal to
If π¦ = βsin π₯ + π¦, then ππ¦/ππ₯ is equal to
Differential coefficient of π‘ππβ1 ( β1+π₯ 2β1 / π₯ ) with respect to π‘ππβ1π₯, when π₯ β 0, is
Derivative of cos π₯. cos 2π₯. cos 3π₯ with respect to π₯ is
If π¦ 2 = π₯ π¦, then ππ¦/ππ₯ is equal to
If π₯π¦ = π (π₯βπ¦) , then ππ¦/ππ₯ is equal to
If π(π₯) = (1 + π₯)(1 + π₯ 2)(1 + π₯ 4)(1 + π₯ 8 ) , then the value of πβ²(π₯)
Derivative of (log π₯) log π₯ with respect to π₯ is
If π¦ π₯ = π π¦βπ₯ , then ππ¦/ππ₯ is equal to
If π₯ = ππ₯/π¦, then ππ¦/ππ₯ is equal to
If π¦ = (cos π₯)(cos π₯)cos π₯β¦β¦β¦..β, then ππ¦/ππ₯ is equal to
If π₯π. π₯π = (π₯ + π¦)π+π, then which of the option is correct
If π₯ = ππ₯π {π‘ππβ1(π¦βπ₯2/π₯2)} , then ππ¦/ππ₯ is equal to
If cos π₯/2 cos π₯/22Β cos π₯/23 β¦ β¦ β¦ β¦ cos π₯/2π = π(π₯), then 1/2 tan π₯/2+1/22Β tan π₯/22 + β― +1/22Β tan π₯/2πΒ is equal to
If π¦ = π ππβ1(1βπ₯2/1+π₯2), 0 < π₯ < 1, then ππ¦/ππ₯ is equal to
If π¦ = πππ β1Β (2π₯/1+π₯2), β1 < π₯ < 1, then ππ¦/ππ₯ is equal to
Differential coefficient of πππ‘β1[β1+sin π₯ +β1βsin π₯ / β1+sin π₯ ββ1βsin π₯], 0 < π₯ <π/2 is
If π¦ = π ππβ1π₯ + π ππβ1β1 β π₯2, β1 β€ π₯ β€ 1, then ππ¦/ππ₯ is
Differentiate πππ β1[sin π₯+cos π₯ / β2] with respect to π₯ is
Differentiate π‘ππβ1[a cos π₯β b sin π₯ / b cos π₯+ a sin π₯] , βπ2< π₯ <π/2 and π/π tan π₯ > β1 , then ππ¦/ππ₯ is
If β1 β π₯2 + β1 β π¦2 = π(π₯ β π¦), then ππ¦/ππ₯ is equal to
Derivative of π ππβ1[1/βπ₯+1] with respect to π₯ is
If y = tππβ1(sec π₯ β tan π₯) ,then ππ¦/ππ₯ is equal to
If π ππβ1π₯ + π ππβ1π¦ =π/2, then ππ¦/ππ₯ is equal to
π/ππ₯ [π ππ2πππ‘β1 {β1βπ₯/1+π₯}] is equal to
π/ππ₯[π ππβ1(π₯β1 β π₯ β βπ₯ β1 β π₯2] is equal to
If π¦ = π‘ππβ1Β [log(π/π₯2 ) / log(ππ₯2)] + π‘ππβ1Β [3+2 πππ π₯ / 1β6 πππ π₯ ] , then π2π¦ / ππ₯2Β is equal to
If π¦ = [π‘ππβ1Β 1/1+π₯+π₯2 + π‘ππβ1Β 1/π₯2+3π₯+3+ π‘ππβ1Β 1/π₯2+5π₯+7+ β― + π’ππ‘π π π‘ππππ ], then π¦β²(0) is equal to