Question

Derivate by first principal


F(x)= √tanx​

Answer 1

Answer:

Step-by-step explanation:

f(x) = tan √x

Then f'(x) = lim.h tends to 0. [f(x+h) - f(x)]/h

= lim.h tends to 0 [tan√(x+h) - tan√x]/h

=lim.h tends to 0 [{sin√(x+h)/cos√(x+h)} - {sin√x/cos√x]/h

=Lim h tends to 0[{sin√(x+h)cos√x-cos√(x+h).sin √x}/cos√(x+h).cos√x] / h

=Lim h tends to 0 [sin {√(x+h) - √x} / cos√(x+h).cosx] / h

=Lim h tends to 0 [sin √x{(√(1 + h/h/x)^2x ) -1} /cos √(x+h).cos√x]. /h

=Lim h tends to 0[ sin √x{1 + 1/2. h/x + (1/2)(1/2 -1)(h/x)^2+……… -1}]/h.cos√(x+h).cos√x.

=Lim h tends to 0[ sin{h/2√x + terms containing higher powers of h/x}]/(h/2√x).2√x.cos√(x+h).cos√x

=Lim h tends to 0[ sin{(h/2√x) + terms containing higher powers of h/x}/(h/2√x).2√x.cos√(x+h).cos√x

=1/{2√x.cos^2 √x} +c

=(sex^2 √x)/(2√x) +c.

Answer 2

HEYA MATE✌️✌️✌️

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solution-:

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