use the rules of differentiation to compute f'(x)

Question

Answer 1

solution to the problem (a)

is

Comments

Thanks a lot :)-

Answer 2

a)f(x) = e^2x/cos(x^3-x)

f(x) = e^2xsec(x^3 -x)

f¹(x) = d/dx [ e^2xsec(x^3 -x)]

Apply d/dx(uv) formula

= e^2x sec(x^3-x) tan(x^3-x) d/dx(x^3-x) +sec(x^3-x) 2e^2x

= e^2xsec(x^3-x)tan(x^3-x) 3x^2 + sec(x^3-x)2e^2x

f¹(x) = e^2xsec(x^3-x) [tan(x^3-x)3x^2+2] .

Answer 3

b) f(x) = (sin^-1x)^3/2

f¹(x) = d/dx ((sin^-1x)^3/2)

       = 3/2(sin^-1x)^(3/2-1) d/dx (sin^-1x)        [apply chain rule and d/dx x^n = nx^n-1]     

       = 3/2(sin^-1x)^1/2 (1/√(1 - x^2)               [d/dx (sin^-1x) =1/√(1 - x^2)]

 

Answer 4

f(x) = log (x^2/(x+2))

f¹(x) = d/dx log (x^2/(x+2))

         = 1/(x^2/(x+2) (d/dx ((x^2/(x+2))

         =  1/(x^2/(x+2)   d/dx (1/x + 2/x^2)                        

f¹(x) = 1/(x^2/(x+2)( logx + (-2/x))

 

Answer 5

d)f(x) = xsinxcosx

f¹(x) = xsinx(-sinx ) + cosx [d/dx(xsinx)]           [d/dx (uv) = ud/dx(v) +vd/dx(u)]

        = xsinx (-sinx) + cosx (xcosx+sinx)

        = -xsin²x +xcos²x+cosxsinx

f¹(x)  = - xsin²x + xcos²x+cosxsinx.          

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