Question

Derivative of (tanx)^(cosx)?

Derivative of (tanx)^(cosx)?

Answer

Let y = [tan(x)]^[cos(x)]. Then, by taking the natural logarithm of both sides yields:

ln(y) = ln{[tan(x)]^[cos(x)]}

==> ln(y) = cos(x)*ln[tan(x)].

By implicit differentiation:

1/y * dy/dx = -sin(x)*ln[tan(x)] + cos(x)*sec^2(x)/tan(x)

==> 1/y * dy/dx = sec(x) - sin(x)*ln[tan(x)]

==> dy/dx = y * {sec(x) - sin(x)*ln[tan(x)]}

==> dy/dx = [tan(x)]^[cos(x)] * {sec(x) - sin(x)*ln[tan(x)]}, since y = [tan(x)]^[cos(x)].

I hope this helps!

Dec 9 at 20:15

y=(tanx)^(cosx)

Take logs base e(i.e. ln)

lny = ln[(tanx)^(cosx)] = coxln(tanx)

Differentiate implicitly

Dec 10 at 0:1

See:

y=(tanx)^(cosx)

y=e^ln(tanx)^(cosx)

y=e^(cosx)ln(tanx)

y '=e^(cosx)ln(tanx).[(cosx)ln(tanx)]'

y '=e^(cosx)ln(tanx).[(cosx) '.ln(tanx)+(cosx)ln(tanx)']

y '=e^(cosx)ln(tanx).[(-sinx) .ln(tanx)+(cosx)(tanx)'/tanx]

y '=e^(cosx)ln(tanx).[(-sinx) .ln(tanx)+(cosx).sec^2(x)/tanx]

y '=e^(cosx)ln(tanx).[(-sinx) .ln(tanx)+sec(x)/tanx]

y '=(tanx)^(cosx).[(-sinx) .ln(tanx)+sec(x)/tanx]

Dec 10 at 4:10

Let y = (tan x)^(cos x)

Then we seek dy/dx. Taking the natural logarithm of both sides, we get

ln y = ln (tan(x) ^ cos(x))

ln y = cos(x) * ln tan(x)

Taking the derivative, d/dx, of both sides, we get

d (ln y) / dx = d(cos x)/dx * ln tan x + cos(x) * d( ln tan x) / dx

(1/y) dy/dx = -sin(x) * ln tan(x) + cos(x) * (1/tan(x)) * (-sec2(x))

(1/y) dy/dx = -sin(x) * ln tan(x) - cos(x) * (cos(x) / sin(x)) * (1/cos2(x))

(1/y) dy/dx = -sin(x) * ln tan(x) - 1/sin(x)

dy/dx = (-sin(x) * ln tan(x) - 1/sin(x))y

dy/dx = (-sin(x) * ln tan(x) - 1/sin(x)) * tan(x) ^ cos(x)

Dec 10 at 8:42