Don't know how to start this..

use chain rule

y=e^(−4tan(x^.5))

dy/dx = e^(−4tan√x) * [-4 Sec² √x] * 1/2√x

dy/dx = -2 e^(−4tan√x) * [Sec² √x] * 1/√x

Use the chain rule:

You have:

f(x) = exp(g(x))

-->

f'(x) = g'(x)exp(g(x))

g(x) = -4tan(x^.5)

--> now use chain rule again

g(x) = -4tan(h(x))

g'(x) = -4h'(x)secÂ²(h(x))

Now h(x) = x^.5

h'(x) = .5 * x^(.5 - 1) = .5x^-.5

--> put that ALL together and you get:

g'(x) = -4*.5x^(-.5) * secÂ²(x^.5)

f'(x) = -2x^(-.5)secÂ²(x^.5)exp(-4tan(x^.5))

or

f'(x) = -2secÂ²(âx)exp(-4tan(âx)) / âx

Make it a little easier on yourself by rewriting the function as e^u, where u = - 4 tan(x^0.5)

Then: Derivative = e^u * du/dx (by chain rule)

Now, finding du/dx: du/dx = - 4 sec^2(x^0.5) * 0.5x^(- 0.5) (applying the chain rule to u)

Putting everything together:

Derivative = - 2x^(- 0.5)*[sec(x^0.5)]^2 * e^(- 4 tan(x^0.5))

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## Answers & Comments

## Verified answer

use chain rule

y=e^(−4tan(x^.5))

dy/dx = e^(−4tan√x) * [-4 Sec² √x] * 1/2√x

dy/dx = -2 e^(−4tan√x) * [Sec² √x] * 1/√x

Use the chain rule:

You have:

f(x) = exp(g(x))

-->

f'(x) = g'(x)exp(g(x))

g(x) = -4tan(x^.5)

--> now use chain rule again

g(x) = -4tan(h(x))

-->

g'(x) = -4h'(x)secÂ²(h(x))

Now h(x) = x^.5

-->

h'(x) = .5 * x^(.5 - 1) = .5x^-.5

--> put that ALL together and you get:

g'(x) = -4*.5x^(-.5) * secÂ²(x^.5)

-->

f'(x) = -2x^(-.5)secÂ²(x^.5)exp(-4tan(x^.5))

or

f'(x) = -2secÂ²(âx)exp(-4tan(âx)) / âx

Make it a little easier on yourself by rewriting the function as e^u, where u = - 4 tan(x^0.5)

Then: Derivative = e^u * du/dx (by chain rule)

Now, finding du/dx: du/dx = - 4 sec^2(x^0.5) * 0.5x^(- 0.5) (applying the chain rule to u)

Putting everything together:

Derivative = - 2x^(- 0.5)*[sec(x^0.5)]^2 * e^(- 4 tan(x^0.5))