Diferentiate y=5^x*sqrt[5^x-(sinx)^2]

Expert Answers
hala718 eNotes educator| Certified Educator

y= 5^x * sqrt(5^x- (sinx)^2]

Let y= u*v such that:

u= 5^x  ==> u' = 5^x* ln 5

v= sqrt[5^x - (sinx)^2 ]

==> v' = (5^x*lnx - 2cosx)*(5^x - (sinx)^2]^-1/2

== v' = [(5^x)*ln5 - 2cosx]/sqrt[(5^x- sinx)^2]

Then,

y' = u'v + uv'

    = (5^x)*[(5^x)*ln5 - 2cosx]/sqrt[5^x-(sinx)^2]+ (5^X)*ln5*sqrt[5^x - (sinx)^2]

giorgiana1976 | Student

We notice that the function is a product, so we'll apply first the product rule.

(u*v)' = u'*v + u*v'

u = 5^x => u' = 5^x*ln5

v = sqrt[5^x-(sinx)^2]

To differentiate the function v, we'll use the chain rule.

v' = {1/2sqrt[5^x-(sinx)^2]}*[5^x-(sinx)^2]'

[5^x-(sinx)^2]' = 5^x*ln 5 - 2sinx*cosx

v' = (5^x*ln 5 - 2sinx*cosx)/2sqrt[5^x-(sinx)^2]

y' = 5^x*ln5*sqrt[5^x-(sinx)^2] + 5^x*(5^x*ln 5 - 2sinx*cosx)/2sqrt[5^x-(sinx)^2]

y' = {5^x*2*ln5*[5^x-(sinx)^2] + 5^x*(5^x*ln 5 - 2sinx*cosx)}/2sqrt[5^x-(sinx)^2]

y' = (5^2x*ln25-5^x*ln25*(sinx)^2+5^2x*ln5-5^x*2sinx*cosx)/2sqrt[5^x-(sinx)^2]

y' = [5^2x*ln 5-5^x*sinx(ln25*sinx-2cosx)]/2sqrt[5^x-(sinx)^2]

neela | Student

To differentiate y = 5^x* sqrt{5^x-(sinx)^2]

Solution:

We know  {u(v(x)}' = (du/dv){dv/dx)

{(u(x)v(x) }' = u'(x)*v(x) + u(x)v'(x).

d/dx(a^x) = (a^x)lna

y' = {(5^x)*sqrt[5^x-sinx)^2]}' = (5^x)'*sqrt[5^x-sinx)^2] +^x*{sqrt[5^x - (sinx)^2]'

= ln5(5^x){sqrt[5^x-(sinx)^2]} +5^x*(1/2){5^x-(sinx)^2}^(1/2_1) * {5^x - (sinx)^2}'

= ln5(5^x){ sqrt{5^x-(sinx)^2} +(1/2){5^x-(sinx)^2}(-1/2) * {ln5(5^x) - 2sinx*cosx}

={ 2ln5(5^x) [5^x-(sinx)^2] +[ln5(5^x) +2sinxcosx]}/sqrt(5^2-(sinx)^2}