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)sec²(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) * sec²(x^.5)
f'(x) = -2x^(-.5)sec²(x^.5)exp(-4tan(x^.5))
or
f'(x) = -2sec²(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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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)sec²(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) * sec²(x^.5)
-->
f'(x) = -2x^(-.5)sec²(x^.5)exp(-4tan(x^.5))
or
f'(x) = -2sec²(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))