Find f’(x) if f(x) = 3^g(x) where g(x) = LOG_3(x^2 - √x + 4/x + 10*LOG_e(x))

if f(x) = 3^g(x)

then f'(x) = (ln3)(3^(gx)) (g ' (x) )

g(x) = ???? , do you mean 3 to be the base of LOG
or are we doing ln ? or what
Once you have decided, continue as below ...
so g ' (x) = ....

insert that result above.

the 3 is the base so log(base3)(...log(base/e)(x)

if g(u) = log_3(u)

f(x) = 3^g(u) = 3^log_3(u) = u
so now it's easy:

f(x) = x^2 - √x + 4/x + 10*ln(x)
f'(x) = 2x - 1/(2√x) - 4/x^2 + 10/x

To find f'(x) if f(x) = 3^g(x), we can apply the chain rule. The chain rule states that if we have a composite function, f(g(x)), the derivative can be found by taking the derivative of the outer function, f'(g(x)), multiplied by the derivative of the inner function, g'(x).

Let's find g'(x) first. We have g(x) = LOG_3(x^2 - √x + 4/x + 10*LOG_e(x)). This is a composition of functions, so we need to find the derivative using the chain rule.

The derivative of LOG_3(u) with respect to u is 1 / (u * ln(3)). Applying the chain rule, we have:

g'(x) = 1 / (x^2 - √x + 4/x + 10*LOG_e(x)) * (2x - (1 / (2√x)) - (4 / x^2) + 10 / x)
= (2x - (1 / (2√x)) - (4 / x^2) + 10 / x) / (x^2 - √x + 4/x + 10*LOG_e(x))

Now, let's find f'(x) using the chain rule:

f'(x) = d(3^g(x)) / dx
= d(3^g(x))/dg * dg/dx
= (3^g(x)) * g'(x) * ln(3)
= 3^g(x) * g'(x) * ln(3)

Substituting g(x) and g'(x) into the equation, we have:

f'(x) = 3^(LOG_3(x^2 - √x + 4/x + 10*LOG_e(x))) * ((2x - (1 / (2√x)) - (4 / x^2) + 10 / x) / (x^2 - √x + 4/x + 10*LOG_e(x))) * ln(3)

Simplifying further is possible, but this is the general expression for f'(x) using the given functions g(x) and f(x).