Homework Help

If y= 3^ (sin x), what is dy/dx?  

user profile pic

timw996 | Student, Grade 10 | (Level 1) Salutatorian

Posted December 3, 2010 at 1:51 AM via web

dislike 0 like

If y= 3^ (sin x), what is dy/dx?

 

3 Answers | Add Yours

user profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted December 3, 2010 at 1:58 AM (Answer #1)

dislike 0 like

To determine the derivative of the given composed function, we'll differentiate both sides:

dy= [3^ (sin x)]'dx

We'll apply chain rule:

dy = 3^ (sin x)*ln 3*(sin x)'dx

But (sin x)' = cos x

dy = 3^ (sin x)*ln 3*(cos x) dx

We'll divide by dx both sides:

dy/dx = (ln 3)*[3^ (sin x)]*(cos x)

user profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted December 3, 2010 at 2:12 AM (Answer #2)

dislike 0 like

y = 3^(sinx).

To find dy/dx.

We know that y = a^f(x) has the differential coefficient , dy/dx = {a^f(x)}' = {(lna)a^f(x)} f'(x).

Therefore  a = 3, f(x) = sinx , f'(x) = (sinx)' = cosx..

Therefore dy/dx = {3^(sinx)}'

dy/dx = {(ln3) 3^(sinx) }(sinx)'

dy/dx = {(ln3)3^(sinx)} cosx.

Therefore the differential coefficient of 3^(sinx) is (ln3^(sinx))(cosx).

user profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted December 3, 2010 at 1:54 AM (Answer #3)

dislike -1 like

It is given that y = 3^ (sin x)

We take the natural logarithm of both the sides and use the relation ln a^b = b*ln a.

=> ln y = ln (3^sin x)

=> ln y = (sin x) (ln 3)

Now differentiate both the sides with respect to x

=> (1/y) (dy/dx) = (ln 3) cos x

=> dy/dx = (ln 3)*y*cos x

=> dy /dx = 3^ (sin x)*(cos x)* ln 3

Therefore if y= 3^ (sin x), dy /dx = 3^ (sin x)*(cos x)* ln 3.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes