differentiate f(x)=Tan^3(e^2x +1)
Let v(x) = e^2x + 1
dv/dx= 2 e^2x
Let u(v) = tan^3 v
f(x) = u[(v(x)]
df/dx = du/dv * dv/dx
For du/dv, let f(g) = g^3 and
g(v) = tan v
u = f{g(v)}
du/dv = df/dg * dg/dv = 3 g^2 * sec^2(v)
= 3 tan^2v*sec^2(v)
df/dx= 2e^(2x)*3tan^2v*sec^2(v)
Substitute e^2x +1 for v for the derivative imn terms of x.
To differentiate the function f(x) = tan^3(e^2x + 1), we will apply the chain rule. The chain rule states that if we have a composite function, such as f(g(x)), where f(u) and g(x) are both differentiable functions, then the derivative of f(g(x)) with respect to x can be found by multiplying the derivative of f(u) with respect to u by the derivative of g(x) with respect to x.
Let's break down the function f(x) = tan^3(e^2x + 1) into its composite parts:
g(x) = e^2x + 1 (inner function)
f(u) = u^3 (outer function)
Now, let's find the derivatives of g(x) and f(u) separately:
Derivative of g(x) = d/dx (e^2x + 1)
The derivative of e^2x is 2e^2x (using the chain rule)
The derivative of 1 is 0
Therefore, the derivative of g(x) = 2e^2x
Derivative of f(u) = d/du (u^3)
Using the power rule, the derivative of u^n with respect to u is n*u^(n-1)
Therefore, the derivative of f(u) = 3u^2
Now, we can apply the chain rule to find the derivative of f(x) = tan^3(e^2x + 1):
d/dx [f(g(x))] = f'(g(x)) * g'(x)
= (3(g(x))^2) * (2e^2x)
= 3(2e^2x) * (e^2x + 1)^2
Therefore, the derivative of f(x) = tan^3(e^2x + 1) is 6e^2x * (e^2x + 1)^2.