How do you differentiate #y=e^(xsecx)+e^5#?

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Answer 1 · 2017-10-22T07:46:57

#(dy)/(dx)=[1+xtanx]secxe^(xsecx)#

#y=e^(xsecx)+e^5#

we want

#(dy)/(dx)=d/(dx)(e^(xsecx))+d/(dx)(e^5)#

now

#d/(dx)(e^5)=0#

since#" "e^5# is a constant

so that leaves us with

#d/(dx)(e^(xsecx))#

now by the chain rule we have the result

#d/(dx)(e^(f(x)))=f'(x)e^(f(x))#

but we also have he product rule to use

#d/(dx)(uv)=v(du)/(dx)+u(dv)/(dx)#

so putting all this together

#(dy)/(dx)=d/(dx)(e^(xsecx))+d/(dx)(e^5)#

#=>(dy)/(dx)=[secxd/(dx)(x)+xd/(dx)(secx)]e^(xsecx)+0#

#:.(dy)/(dx)=[secx+xsecxtanx]e^(xsecx)#

#(dy)/(dx)=[1+xtanx]secxe^(xsecx)#