What is the derivative of f(x)=5^x log6(x)?
Exponential rule:
d(a^x)/dx = ln(a) a^x
For logarithms to arbitrary base:
d(loga(x))/dx
=d(ln(x)/ln(a))/dx
=ln(a)/x
Finally, the product rule:
d(uv)/dx = u dv/dx + v du/dx
set
u=5^x,
v=log6(x)
=ln(x)/ln(6)
Can you take it from here?
For logarithms to arbitrary base:
d(loga(x))/dx
=d(ln(x)/ln(a))/dx
=1/(ln(a)*x)
To find the derivative of the function f(x) = 5^x log6(x), we can use the product rule.
The product rule states that if we have a product of two functions, u(x) and v(x), the derivative of their product is given by the formula:
(f * g)' = f' * g + f * g'
Let's apply this rule to our function:
f(x) = 5^x log6(x)
We can rewrite this as:
f(x) = (5^x) * (log6(x))
Now, let's define our functions:
f(x) = u(x) * v(x)
where
u(x) = 5^x
v(x) = log6(x)
Now, let's find the derivatives of u(x) and v(x):
The derivative of u(x) = 5^x can be found using the chain rule. The chain rule states that if we have a composite function y = u(v(x)), then the derivative is given by:
dy/dx = u'(v(x)) * v'(x)
In our case, u(x) = 5^x can be written as:
y = u(v) = 5^v
where v = x.
Thus, the derivative of u(x) = 5^x is:
u'(x) = 5^x * log(5)
Now, let's find the derivative of v(x) = log6(x). The derivative of the natural logarithm function, log(x), is 1/x. However, in this case, we have log6(x), which means we need to apply the chain rule.
Using the chain rule, we have:
v'(x) = (1/x) * (1/ln(6))
Now, let's substitute the derivatives we found into the product rule:
f'(x) = u'(x) * v(x) + u(x) * v'(x)
Substituting the values we found:
f'(x) = (5^x * log(5)) * log6(x) + (5^x) * (1/x) * (1/ln(6))
And there you have it! The derivative of f(x) = 5^x log6(x) is given by the expression above.