find the derivative of y = (2+sinx)^x?

Answer 1

So, let's take the natural log of both sides. We get

ln(y)=ln(2+sin(x))^x

ln(y)=x*ln(2+sin(x))

Differentiate both sides with respect to x, and you have:

(1/y)(dy/dx)=ln(2+sin(x))+x(cos(x)/(2+sin(x))

Therefore, just multiply both sides by y, and

dy/dx = y[ln(2+sin(x)) + x(cos(x)/(2+sin(x))]

Where y = (2+sin(x))^x

Yes, the answer is a bit ugly, and I'm sure your teacher will probably let you leave a y in the answer as long as you make clear that y = (2+sin(x))^x.

=)

Answer 2

ok so this is a complex derivative, one that you have to know a little rule to get.
you have to know how do logarithmic differentiation.

you have to start out by taking the the natural log of both sides,

ln y = ln[(2+sinx)^x]

ln y = xln(2+sinx)

then differentiate implicitly with respect to x so

(1/y) y' = x[1/(2+sinx)] (cosx) + ln(2+sinx)

so then solve for then solve for y' and we get
y' = y {x[1/(2+sinx)] (cosx) + ln(2+sinx)}

so in the equation for y' we have a y, but we were given y intially so y = (2+sinx)^x, so just plug in y,

so the final answer is
y' = [(2+sinx)^x ] {x[1/(2+sinx)] (cosx) + ln(2+sinx)}

yea this one is a long nasty derivative

email for any questions

Answer 3

use logarhytmic differentiation:

ln y = ln (2+sinx)^x

ln y = x * ln (2+sin x)

differentiate both sides with respect to x

(1/y)*y' = (1*ln(2+sinx)+x*(cosx))

mutiply both sides by y to get

y' = y*[ln(2+sinx+xcosx)]

now substitute for y

y'=(2+sinx)^x * [ln(2+sinx+xcosx)]

Answer 4

you first have to take the ln of both sides to move x out of the exponent. then it's
ln y=x(ln 2+sinx)
then take the derivitive and solve for y'

Answer 5

make:

y = (2 + sinx)^z, then
dy = (dy/dx)*dx + (dy/dz)*dz
= z*(2 + sinx)^(z-1) *cosx *dx + (2 + sinx)^z*ln(2+sinx) dz

now going back z=x you get:

dy/dx = x*(2 + sinx)^(x-1)*cosx + (2 + sinx)^x*ln(2+sinx)
= (2 + sinx)^(x-1)*[ x*cosx + (2 + sinx)*ln(2 + sinx)].

Answer 6

huh?