derivative of tan cube root x
Answers
Answer:
Derivative of tanx−−−−√3 by First Principle is given by:
=>f′(x)=limh→0 tan(x+h)√3−tanx√3h
=>limh→0 sin(x+h)cos(x+h)√3−sinxcosx√3h
=>limh→0 sin(x+h)cos x√3−sinx cos(x+h)√3hcos(x+h)cosx√3
=>limh→01cos(x+h)cosx√3. limh→0 sin(x+h)cos x√3−sinx cos(x+h)√3h
=>1cos(x+0)cosx√3. limh→0 sin(x+h)cos x√3−sinx cos(x+h)√3h
=>1cos2x√3. limh→0 sin(x+h)cos x√3−sinx cos(x+h)√3h×[(sin(x+h)cos x)23+(sinx cos(x+h))23+sin(x+h)cosx. cox(x+h)sinx√3][(sin(x+h)cos x)23+(sinx cos(x+h))23+sin(x+h)cosx. cox(x+h)sinx√3]
=>1cos2x√3. limh→0 sin(x+h)cos x−sinx cos(x+h)h[(sin(x+h)cos x)23+(sinx cos(x+h))23+sin(x+h)cosx. cox(x+h)sinx√3] [a3−b3=(a−b)(a2+b2+ab)]
=>1cos2x√3. limh→0 sin(x+h−x)h[(sin(x+h)cos x)23+(sinx cos(x+h))23+sin(x+h)cosx. cox(x+h)sinx√3] [sin(a−b)=sina.cosb−cosa.sinb]
=>1cos2x√3. limh→0 sinhh. limh→01[(sin(x+h)cos x)23+(sinx cos(x+h))23+sin(x+h)cosx. cox(x+h)sinx√3]
=>1cos23x.1.1[(sin(x+0)cos x)23+(sinx cos(x+0))23+sin(x+0)cosx. cox(x+0)sinx√3]
=>1cos23x.1.1[(sinxcos x)23+(sinx cosx)23+sinx.cosx. cosx.sinx√3]
=>1cos23x.1.1[(sinxcos x)23+(sinx cosx)23+(sinxcosx)23]
=>1cos23x.13(sinxcos x)23
=>131cos43x.sin23x
=>131cos2x(cos43x.sin23xcos2x)
=>13sec2x(sin23xcos23x)
=>13.1tan23x.sec2x
Step-by-step explanation: