Question

Derivate by first principal

f(x). = secx​

Answer 1

Heya mate
The answer of ur question is





'(x)=limh→0f(x+h)−f(x)h

        =limh→0sec(x+h)−sec(x)h

        =limh→01cos(x+h)−1cos(x)h

        =limh→0cosx−cos(x+h)cos(x+h)cos(x)h

        =limh→0cosx−cos(x+h)hcos(x+h)cos(x)

        =limh→0cosx−(cosxcosh−sinxsinh)hcos(x+h)cos(x)

        =limh→0cosx−cosxcosh+sinxsinhhcos(x+h)cos(x)

        =limh→0cosx(1−cosh)+sinxsinhhcos(x+h)cos(x)

        =limh→0cosx(1−cosh)hcos(x+h)cos(x)+sinxsinhhcos(x+h)cos(x)

        =limh→0(1−cosh)hcos(x+h)+tanxsinhhcos(x+h)

        =limh→01−coshh⋅sec(x+h)+sinhh⋅tanxsec(x+h)

Then we use two standard calculus limits:

limθ→01−cosθθ=0 and limθ→0sinθθ=1

Which gives us:

f'(x)=0⋅sec(x)+1⋅tanxsec(x)
        =tanxsecx

Answer 2

HEY mate


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