Question

If Y=e^ax•COS^3 X•SIN^2 X find dy/dx

Answer 1

Let F_= e^ax

Answer 2

G_=COS^3X

Answer 3

& H_=SIN^2X :

Answer 4

Y'=(F•G•H)'

Answer 5

=[F•(G•H)]'

Answer 6

=F'•(GH)+F•(GH)'

Answer 7

→ Y'=F'GH+FG'H+FGH'

Answer 8

F'=ae^ax,

Answer 9

G'=-3SINX•COS^2X ,

Answer 10

H'= 2SINX•COSX

Answer 11

Therefore: y'=e^ax•SINX•COS^2X•(a SINX•COSX-3SIN^2X+2COS^2X)

Answer 12

y=e^(ax)·cos³x·sin²x  Take the natural logarithm of the absolute values of both sides.  Since ln(e)=1, so that the equation can be written as follows: ln|y|=ax+3·ln|cosx|+2·ln|sinx| ··· Eq.1

Since (d/dy)·ln|y|=1/y, so that, (d/dx)·ln|y|=((d/dy)·ln|y|)·(dy/dx)=y'/y, and also (ln(cosx))'=-sinx/cosx,    (ln(sinx))'=cosx/sinx, both sides of Eq.1 can be differentiated with respect to x and simplified as follows: y'/y=1a-3sinx/cosx+2cosx/sinx=(a·cosx·sinx-3sin²x+2cos²x)/(cosx·sinx)

Therefore, y'=(e^(ax)·cos³x·sin²x)·((a·cosx·sinx-3sin²x+2cos²x)/(cosx·sinx))=e^(ax)·cos²x·sinx·(a·cosx·sinx-3sin²x+2cos²x)

Answer: dy/dx=e^(ax)·cos²x·sinx·(a·cosx·sinx+2cos²x-3sin²x) 

Answer 13

y=e^(ax)·cos³x·sin²x  Use the product rule for calculating the derivatives of product-form-functions: if y=u·v·w and u,v,w are functions of x, y'=(u)'·v·w+u·(v)'·w+u·v·(w)'

Therfore, y'=(e^(ax))'·cos³x·sin²x+e^(ax)·(cos³x)'·sin²x+e^ax·cos³x·(sin²x)' ··· Eq.1

Here, use the chain rule for the differentiation of composite functions: if y=f(u) and u=f(x), dy/dx=dy/du·du/dx 

Therfore Eq.1 can be restated: y=(e^(ax)·1a)·cos³x·sin²x+e^(ax)·(3cos²x·(-sinx))·sin²x+e^(ax)·cos³x·(2sinx·cosx)=e^(ax)·cos²x·sinx·(a·cosx·sinx-3sin²x+2cos²x)

Answer: dy/dx=e^(ax)·cos²xsinx·(a·cosx·sinx+2cos²x-3sin²x)