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

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

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

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

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

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