find the derivative of the function using chain rule

Question

a) y = e^axcos(bx)

b) f(x) = ((1+ax)/(1+e^bx))¹⁰

c) g(x) = √e^sin(x) + (x + cos⁴(2x))³

Answer 1

y = e^axcos(bx)

y' = e^(ax) * -b*sin(bx) + a*e^(ax)*cos(bx)

e^(ax) (a*cos(bx) - b*sin(bx))

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f(x) = ((1+ax)/(1+e^bx))¹⁰

^{y' = { a*(1 + exp(bx))^10 - 10*b*(1 + ax)*(1 + exp(bx))^9*exp(bx)} / (1 + exp(bx))^20}

^{factors out (1 + exp(bx))^9 from the numerator which makes the exponent in the denominator 11}

{ a * (1 + exp(bx) - 10*b* (1 + ax)*exp(bx))}/ (1 + exp(bx))^11

{a + a*exp(bx) - 10b*exp(bx) - 10ax*exp(bx)}/(1 + exp(bx))^11

^{==============================================================================}

g(x) = √e^sin(x) + (x + cos⁴(2x))³

^{y = [exp(sin x)]^(1/2) + [ x + (cos(2x))^4]^3}

takes log of both sides

^{ln y = (1/2) (sinx) + 3 * ln[ x + (cos(2x))^4]}

^{y'/y = {(1/2)(cosx) + 3 (1 + 4(cos(2x))^3*-2sin(2x))}/ [x + ((cos(2x))^4)]}

= {(1/2)(cosx) + 3 -24(cos(2x))^3*sin(2x)}^{/ [x + ((cos(2x))^4)]}

multiplies both sides by y and substitutes y=f(x)

y' = ^{[exp(sin x)]^(1/2) + [ x + (cos(2x))^4]^3*}{(1/2)(cosx) + 3 -24(cos(2x))^3*sin(2x)}^{/ [x + ((cos(2x))^4)]}