Please help! I'm do know how to solve this question. Please anyone please solve and show the steps too, thanks.

Question

Differentiate each of the following.

a) y = e^πx b) y = 8^4x

c) y = (e^{x^2}) / (x) d) y = x2^3x+2

Answer 1

y = e^pix

y'(x) = pi(e^pix)

use the chain rule

y=e^u with u=pix, du = (pi)dx

y'(u) = e^u)du = e^pix)(pi)dx)

y'(x) = pi(e^pix)

or take natural logs and differentiate, then substitute back for the value of y in terms of x

y= e^pix

lny = pix

(1/y)y' = pi

multiply both sides by y

y' =ypi

substitute the value of y in terms of x

y'= e^(pix)pi

y' = pi[e^(pix)]

y= 8^4x

take logs to the base 8 to both sides

log₈y = 4x

convert to natural logs

lny/ln8 = 4x

lny = 4x/ln8

differentiate both sides

(1/y)y' = 4/ln8

substitute 8^4x for y

y' =(4/ln8)(8^4x)

y'= about 1.924(8x^4x)

y=(e^(x^2))/x

xy = e^(x^2)

lnxy = x^2

(1/xy)y' = 2x

y' = 2yx^2

y' = (2x^2)(e^(x^2)

y=x(2^(3x+2)

y/x = 2^(3x +2)

log₂(y/x) = 3x+2

convert to natural log

ln(y/x)/ln2 = 3x+2

ln(y/x) = (3x+2)/ln2

lny - lnx = 3x/ln2 + 2/ln2

lny = 3x/ln2 +2/ln2 + lnx

differentiate both sides

(1/y)y' = (3/ln2) + 1/x

y' = 3y/ln2 + y/x

y' =y[(3/2)/ln2 + 1/x)

substitute x(2x^(2x+3)) for y

y' = (x(2x^(2x+3)[3/2ln2 + 1/x]

these calculations are endlessly tedious. but hopefully no error, but if there's an error somewhere, this solution still conveys the basic method.