Studdy Solution
STEP 1
What is this asking?
We need to find the derivative of a function where both the base and the exponent have x in them!
Watch out!
We can't use the power rule directly here since the exponent isn't constant.
Logarithmic differentiation is our friend!
STEP 2
1. Set up for Logarithmic Differentiation
2. Compute the Derivative of Both Sides
3. Isolate and Simplify
STEP 3
Let's take the natural log of both sides of f(x)=(ln(x))sec(x).
This gives us:
ln(f(x))=ln((ln(x))sec(x))
Why are we doing this?
Because logarithms have a cool property that lets us bring down exponents, which will make this much easier to differentiate!
STEP 4
Using the power rule for logarithms, we can bring the sec(x) down:
ln(f(x))=sec(x)⋅ln(ln(x))
Now we have a product, which we know how to differentiate!
STEP 5
Remember the chain rule!
The derivative of ln(f(x)) is f(x)f′(x).
So, differentiating both sides gives us:
f(x)f′(x)=dxd(sec(x)⋅ln(ln(x)))
We're one step closer to finding f′(x)!
STEP 6
We need the product rule to differentiate the right side.
Remember, the product rule says (uv)′=u′v+uv′.
Here, u=sec(x) and v=ln(ln(x)).
So, u′=sec(x)⋅tan(x) and v′=ln(x)1⋅x1=x⋅ln(x)1.
Putting it all together:
f(x)f′(x)=sec(x)⋅tan(x)⋅ln(ln(x))+sec(x)⋅x⋅ln(x)1
Look at us go!
STEP 7
To isolate f′(x), we multiply both sides by f(x), which is (ln(x))sec(x):
f′(x)=(ln(x))sec(x)⋅(sec(x)⋅tan(x)⋅ln(ln(x))+x⋅ln(x)sec(x))
Almost there!
STEP 8
We can factor out sec(x) to make it look even nicer:
f′(x)=(ln(x))sec(x)⋅sec(x)⋅(tan(x)⋅ln(ln(x))+x⋅ln(x)1)
And there we have it!
STEP 9
f′(x)=(ln(x))sec(x)⋅sec(x)⋅(tan(x)⋅ln(ln(x))+x⋅ln(x)1)
