Differentiate #secx# using the definition of differential?

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Answer 1 · 2017-01-26T09:29:06

#d/(dx)secx=secxtanx#

For a function #f(x)#, its differential #(df)/(dx)=Lt_(h->0)(f(x+h)-f(x))/h#

Hence for #secx#, #d/(dx)secx=Lt_(h->0)(sec(x+h)-secx)/h#

= #Lt_(h->0)(1/cos(x+h)-1/cosx)/h#

= #Lt_(h->0)(cosx-cos(x+h))/(hcos(x+h)cosx)#

= #Lt_(h->0)(2sin((x+h+x)/2)sin((x+h-x)/2))/(hcos(x+h)cosx)#

= #Lt_(h->0)(sin(x+h/2)sin(h/2))/(h/2cos(x+h)cosx)#

= #Lt_(h->0)sin(x+h/2)/(cos(x+h)cosx)xxLt_(h->0)(sin(h/2))/(h/2)#xx

= #sinx/(cosxcosx) xx1#

= #1/cosx xxsinx/cosx#

= #secxtanx#