Let f(x)=8(sin(x))^x. Find f′(3).
f′(3)=
First you have to find out what f'(x) is, and then plug 3 into that.
d/dx [8(sin(x))^x] = 8 x [(sin(x))^(x-1)] cos(x)
if x = 3 then
8*3 [ sin^2 (3)] cos(3) ]
I assume x is in radians
If u and v are functions of x, then d/dx u^v is a combination of the power rule and the exponent rule:
d/dx u^v = v u^(v-1) u' + lnu u^v v'
So, d/dx (sinx)^x = x (sinx)^(x-1) * cosx + ln sinx * (sinx)^x * 1
Missed that, use oobleck solution !
To find the derivative of a function, we can use the chain rule. The chain rule states that if we have a composition of functions, then the derivative is the derivative of the outer function times the derivative of the inner function.
In this case, we have f(x) = 8(sin(x))^x. To find f'(3), we need to find the derivative of f(x) and then evaluate it at x = 3.
First, let's find the derivative of f(x) using the chain rule.
Let's define u = 8(sin(x))^x. We can rewrite f(x) in terms of u as f(x) = u.
Now, we need to find the derivative of u with respect to x. To do this, we can take the natural logarithm of both sides of the equation:
ln(f(x)) = ln(u)
Applying the logarithm rule, we can bring down the exponent:
ln(f(x)) = x ln(8(sin(x)))
Next, we can differentiate both sides of the equation with respect to x:
(1/f(x)) * f'(x) = ln(8(sin(x))) + x * [1/(8(sin(x))) * (8cos(x))]
To isolate f'(x), we can multiply both sides of the equation by f(x):
f'(x) = f(x) [ln(8(sin(x))) + x * [1/(8(sin(x))) * (8cos(x))]]
Remember that we defined f(x) = u, so we can substitute u back in:
f'(x) = u [ln(8(sin(x))) + x * [1/(8(sin(x))) * (8cos(x))]]
Now, we can evaluate f'(3) by plugging in x = 3 into the derivative expression:
f'(3) = u [ln(8(sin(3))) + 3 * [1/(8(sin(3))) * (8cos(3))]]
Thus, f'(3) = u [ln(8(sin(3))) + 3 * [1/(8(sin(3))) * (8cos(3))]]