Find the derivative of the following using first principal f(x)=tanx

1.	Find the derivative of the following using first principal f(x)=tanx
Answer» By first principle if f[x]=tanx F'[x]= f[x+h]-f[x]/x+h-x at Lth tends to 0 F'(x) = tan(x+h)-tanx/h By simple trigonometry tan(x+h)=tanx+tanh/1-tanxtanh Thus, f'(x) = tanx+tanh-tanx+tan^2xtanh/(1-tanxtanh) h =tanh(1+tan^2x)/(1-tanxtanh) h But Lt (tanx)/x. =1, 1+tan^2x=sec^2x h~0 Hence f'(x) =sec^2x(1)/(1-tanxtanh) = sec^2x Thanks

Answer»

By first principle if f[x]=tanx

F'[x]= f[x+h]-f[x]/x+h-x at Lth tends to 0

F'(x) = tan(x+h)-tanx/h

By simple trigonometry

tan(x+h)=tanx+tanh/1-tanxtanh

Thus,

f'(x) =

tanx+tanh-tanx+tan^2xtanh/(1-tanxtanh) h

=tanh(1+tan^2x)/(1-tanxtanh) h

But Lt (tanx)/x. =1, 1+tan^2x=sec^2x

h~0

Hence f'(x) =sec^2x(1)/(1-tanxtanh)

= sec^2x

Thanks

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