If y=8^(cos x), what is dy/dx?
We have y = 8^cos x.
We use the chain rule to find the derivative of 8^(cos x)
Now the derivative of a^x is a^x log a and the derivative of cos x is -sin x.
y' = log 8 * 8^cos x * (cos x)'
=> y' = -log 8 * 8^cos x* sin x
Therefore dy/dx = -log 8 * 8^cos x* sin x
y = 8^cosx.
We take the logarithms of both sides:
lny = (cosx)(ln8).
We differentiate both sides with respect to x.
(1/y)(dy/dx) = (ln8)*(cosx)'.
dy/dx = y*(ln8)(-sinx).
dy/dx = (ln8)(8^(cosx))(-sinx)..
Therefore the differential coefficient of y = 8^(cosx) is dy/dx = -(ln8)(8^(cosx))sinx.
To determine the derivative of the given composed function, we'll differentiate both sides:
dy= [8^ (cos x)]'dx
We'll apply chain rule:
dy = 8^ (cos x)*(ln 8)*[(cos x)']dx
But (cos x)' = -sin x
dy = -8^ (cos x)*ln 8*(sin x) dx
We'll divide by dx both sides:
dy/dx = -[8^ (cos x)]*(ln 8)*(sin x)