find the derivative of 9^8^x^2
let y = 9^8^(x^2)
I assume these are staggered exponents.
take ln of both sides
lny = ln(9^8^(x^2))
lny = (8^)x^2))ln9
now differentiate
y'/y = (ln9)(ln8)(8^(x^2))(2x)
y' = dy/dx = 2xy(ln8)(ln9)8^(x^2)
no its 9 to the power 8 to the power x to the power 2
To find the derivative of the function 9^(8^(x^2)), we can use the chain rule. The chain rule states that if we have a function composed of two functions, f(g(x)), then the derivative is given by the product of the derivative of the outer function (f'(g(x))) and the derivative of the inner function (g'(x)).
Let's break down the given function:
f(x) = 9^(8^(x^2))
We can rewrite this function as:
f(x) = 9^(8^u), where u = x^2
Now we have an outer function f(u) = 9^u, and an inner function u(x) = 8^x^2.
To find the derivative, we begin by finding the derivative of the outer function f(u):
f'(u) = d/dx (9^u)
To do this, we can use logarithmic differentiation.
First, we take the natural logarithm of both sides:
ln(f(u)) = ln(9^u)
Applying the logarithmic property ln(a^b) = b*ln(a), we have:
ln(f(u)) = u * ln(9)
Next, we differentiate both sides with respect to u:
d/dx [ln(f(u))] = d/dx (u * ln(9))
To differentiate ln(f(u)), we apply the chain rule:
f'(u) / f(u) = 1 * ln(9)
Now, we can solve for f'(u):
f'(u) = f(u) * ln(9)
Now, we need to find the derivative of the inner function u(x) = 8^x^2:
u'(x) = d/dx (8^x^2)
To do this, we can again use the chain rule.
Let's rewrite the function:
u(x) = 8^(x^2)
Taking the natural logarithm of both sides:
ln(u(x)) = ln(8^(x^2))
Using the logarithmic property, we have:
ln(u(x)) = x^2 * ln(8)
Differentiating both sides with respect to x:
d/dx [ln(u(x))] = d/dx (x^2 * ln(8))
Applying the chain rule to ln(u(x)):
u'(x) / u(x) = 2x * ln(8)
Now, solve for u'(x):
u'(x) = u(x) * 2x * ln(8)
Finally, we can apply the chain rule by multiplying the derivatives of the outer and inner functions:
f'(x) = f'(u) * u'(x)
Substituting the derived expressions:
f'(x) = (f(u) * ln(9)) * (u(x) * 2x * ln(8))
Plugging back the original functions:
f'(x) = (9^(8^u) * ln(9)) * (8^(x^2) * 2x * ln(8))
Therefore, the derivative of 9^(8^(x^2)) is (9^(8^u) * ln(9)) * (8^(x^2) * 2x * ln(8)) where u = x^2.