.done challenging questions of calcus in simple formula.

May 29th, 2016
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done tough questions of calculus good for revision.

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Running head: CALCULUS 1 ASSIGNMENTCALCULUSNAMEINSTITUTIONRunning head: CALCULUS 1 ASSIGNMENTQ .1AConverting it quotient and applying l hospitals rule,Limx2X - -e -xApplying l hospital rule,Lim2xX- -e -xLim 2X-2 =0e -xQ. BINDETERMINATE FORM 0/0Limpi/2- xXPi/2 cot xApplying L hospital ruleLim1Xpi/2 csc2xQ.1DLim sin x-xX0 X3Applying l hospitals rule,Limcos x -1X0 3x 2Limsin xX06xLimcos x = -1/6X06Running head: CALCULUS 1 ASSIGNMENTQ3.Find the derivative of the3 A.(Sin-1x)2 is same as arcsin2xf(x)= arcsin2x + pi^6 +6^xDifferentiate term by term,f(x ) = d/dx(arcsin2x) + d/dx(6^x) + d/dx(pi^6)f(x ) = 2arcsin(x)*d/dx(arcsin x)+ ln(6)*6^x +0f(x ) =2*(1/1-x 2)arcsinx + ln(6)*6^xf(x ) = 2sin-1x/ (1-x2) + ln(6)*6^x3B.F(x) = ln(x 3sinx)Using identity and differentiating,ln(ab) = lna + ln bf(x) = Ln x 3 + ln(sinx)f(x) = 1/x 3*3x 2 + cosx/ sinx, and simplifying,f(x) = 3/ x + cot x3cFACTOR OUT AND SIMPLIFYRunning head: CALCULUS 1 ASSIGNMENTLimx 2-1X-1x+1By l hospital rule, differentiate once, we getLimX-12x = -21Lim2x 2 + x-3 =-2X-1+The function f(x) is continuous at x=-1, since the left hand limit is equal to the right hand limit.Q4.dy/dx=1/3(2x 2 +4)-2/3*4xdy/dx=4/3x*(2x 2+4)at point x=2 then dy/dx=4/32*(2*(2(2) 2 +4)dy/dx=4/32*(2*2+4)= 4/3*2*8dy/dx=(32*2)/3equation of the tagent at the point (2,2) (x,y)y-2/x-2=

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