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Queastion: differentiate [sin (e^x) - cos (Inx)] ^100 .

Use the chain rule.

Let u = sin (e^x) - cos (Inx)

Let v = e^x and w = ln(x), giving dv/dx = e^x and dw/dx = 1/x, then

u = sin(v) - cos(w), and

du/dx = (d(sin(v))/dv)*(dv/dx) - (d(cos(w))/dw)*(dw/dx)

du/dx = cos(v)*e^x + sin(w)*(1/x)

du/dx = (e^x)*cos(e^x) + (1/x)*sin(ln(x))

Finally,

y = u^100, with dy/du = 100u^99

So,

dy/dx = (dy/du)*(du/dx)

dy/dx = 100[sin (e^x) - cos (Inx)]^99*[(e^x)*cos(e^x) + (1/x)*sin(ln(x))]

by Level 11 User (80.5k points)

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