$${\frac{{\mathtt{dy}}}{{\mathtt{dx}}}}$$ 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)
+33614 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
+33614 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
