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$${\frac{{\mathtt{dy}}}{{\mathtt{dx}}}}$$ of $${\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{e}}}^{{\mathtt{y}}} = {{\mathtt{x}}}^{{\mathtt{2}}}$$

andf(x)= $${{\mathtt{sinxcos}}}^{{\mathtt{3}}}$$X in to f'(x)

Guest Sep 18, 2014

Best Answer 

 #2
avatar+26973 
+8

y + ey = x2

Differentiating this with respect to x we get:

dy/dx + ey*dy/dx = 2x   (Use the chain rule for dey/dx: i.e.  dey/dy*dy/dx)

(1 + ey)*dy/dx = 2x

dy/dx = 2x/(1 + ey)

 

f(x) = sin(x)*cos3(x) 

df(x)/dx = sin(x)*dcos3(x)/dx + cos3(x)*dsin(x)/dx

df(x)/dx = sin(x)*3cos2(x)*(-sin(x)) + cos3(x)*cos(x)

df(x)/dx = cos2(x)*(-3sin2(x) + cos2(x))

Now since cos(2x) = cos2(x) - sin2(x) = 1 - 2sin2(x) = 2cos2(x) - 1 we could write

df(x)/dx = (cos(2x) + 1)*(2cos(2x) - 1)/2

Alan  Sep 18, 2014
 #2
avatar+26973 
+8
Best Answer

y + ey = x2

Differentiating this with respect to x we get:

dy/dx + ey*dy/dx = 2x   (Use the chain rule for dey/dx: i.e.  dey/dy*dy/dx)

(1 + ey)*dy/dx = 2x

dy/dx = 2x/(1 + ey)

 

f(x) = sin(x)*cos3(x) 

df(x)/dx = sin(x)*dcos3(x)/dx + cos3(x)*dsin(x)/dx

df(x)/dx = sin(x)*3cos2(x)*(-sin(x)) + cos3(x)*cos(x)

df(x)/dx = cos2(x)*(-3sin2(x) + cos2(x))

Now since cos(2x) = cos2(x) - sin2(x) = 1 - 2sin2(x) = 2cos2(x) - 1 we could write

df(x)/dx = (cos(2x) + 1)*(2cos(2x) - 1)/2

Alan  Sep 18, 2014
 #3
avatar+93351 
+3

Thanks Alan,

I was waiting for you to answer that first one.  

Melody  Sep 18, 2014

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