Differentiation

[SuperTramp]

How do I differentiate y=e^cos(x) ?

[DaveD]

Use the chain rule

let u = cos(x)

so y= e^u

so dy/du = e^u = e^cos(x) and du/dx = -sin(x)

since dy/dx = dy/du * du/dx

then dy/dx = e^cos(x) * sin(x)

= sin(x) e^cos(x)

[pwlldvsbj12]

Actually no its -sin(x)e^cos(x) because:

y = e^cos(x)

ln(y) = cos(x)

then implicit differentiation:

1/y X dy/dx = -sin(x)

dy/dx = y X -sin(x)

but y = e^cos(x)

so dy/dx = e^cos(x) X -sin(x)

= -sin(x)e^cos(x)