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Question

Find the derivative of the following functions:
5secx+4cosx

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Solution

Let f(x)=5secx+4cosx
Thus using first principle,
f(x)=limh0f(x+h)f(x)h
= limh05sec(x+h)+4cos(x+h)[5secx+4cosx]h
= 5limh0[sec(x+h)secx]h+4limh0[cos(x+h)cosx]h
= 5limh01h[1cos(x+h)1cosx]+4limh01h[cos(x+h)cosx]
= 5limh0[cosxcos(x+h)cosxcos(x+h)]+4limh01h[cosxcoshsinxsinhcosx]
= 5cosxlimh01h⎢ ⎢2sin(2x+h2)sin(h2)cos(x+h)⎥ ⎥+4[cosxlimh0(1cosh)hsinxlimh0sinhh]
= 5cosxlimh0⎢ ⎢ ⎢ ⎢ ⎢ ⎢sin(2x+h2)sin(h2)h2cos(x+h)⎥ ⎥ ⎥ ⎥ ⎥ ⎥+4[(cosx)(0)(sinx)1]
= 5cosx⎢ ⎢limh0sin(2x+h2)cos(x+h)limh0sin(h2)h2⎥ ⎥4sinx
= 5cosxsinxcosx14sinx
= 5secxtanx4sinx

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