Calculus: Find g'(3) if g(x)=x2^(h(x)) where h(3)=-2 and h'(3)=5??

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Answer 1 · 2018-05-10T08:15:09

#g'(3)=1/4(1+15ln2)#

We know that,

#(1)color(violet)(d/(dx)(a^x)=a^x*lna#

We have,

#g(x)=x*2^(h(x))#

First, we take

#u=2^(h(x))=2^vwhere,v=h(x).#

#i.e. u=2^v andv=h(x)...toApply(1)#,for #a=2#

#=>(du)/(dv)= 2^vln2and (dv)/(dx)=h'(x)#

#"Using "color(blue)"Chain Rule :"#

#color(blue)((du)/(dx)=(du)/(dv)xx(dv)/(dx))=2^vln2*h'(x)#

#color(red)(d/(dx)(2^(h(x)))=2^(h(x))ln2*h'(x)#

Now,

#g(x)=x*2^(h(x))#

Diff.w.r.t.#x# , #"using "color(blue)"Product Rule : "#

#=>g'(x)=x*color(red)(d/(dx)(2^(h(x))))+2^(h(x))*d/(dx)(x)#

#=>g'(x)=x[color(red)(2^(h(x))ln2*h'(x))]+2^(h(x)) (1)#

#=>g'(x)=2^(h(x))[xln2*h'(x)+1}#

Subst. # x=3# , we get

#g'(3)=2^(h(3))[3ln2*h'(3)+1]#

Given that,

#h(3)=-2 and h'(3)=5#

So,

#g'(3)=2^(-2)[3ln2xx5+1]#

#=>g'(3)=1/4(1+15ln2)#