g'(x)=tan(2/1+x^2)
Let g be the function with first derivative given above and g(1)=5. If f is the function defined by f(x)=ln(g(x)), what is the value of f'(1)?
I know the answer is 0.311, but I need steps as to why. Please and thanks.
Well, I could go into a long and boring explanation about derivatives and trigonometric functions, but where's the fun in that? Let's tackle this problem together in a fun and silly way!
First things first, let's consider the given function g'(x) = tan(2/(1 + x^2)) and g(1) = 5. We want to find the value of f'(1) where f(x) = ln(g(x)).
Step 1: Start with the given information - g(1) = 5. But what does g(1) = 5 mean in real-life terms? Well, let's say g(1) is like having a pet dinosaur named Gary. So, Gary the Dinosaur has a value of 5. Great!
Step 2: Moving on, we know that f(x) = ln(g(x)). In other words, f(x) is like taking the natural log of Gary the Dinosaur. So, f(x) is the logarithm of a pet dinosaur! How cool is that?
Step 3: Now, we want to find f'(1), which is basically asking how fast does the logarithm of Gary the Dinosaur change when x = 1. We want to know the rate at which Gary's log is growing or shrinking.
Step 4: To find f'(1), we need to use the chain rule. You remember the chain rule, right? It's like taking a chain and using it to keep your pet dinosaur from running away. Well, in math terms, the chain rule helps us find the derivative of functions like f(x) which involve other functions, like g(x).
Step 5: Applying the chain rule, we have f'(x) = g'(x) / g(x). It's like the derivative of f(x) is equal to the derivative of g(x) divided by g(x). Fancy, huh?
Step 6: Plugging in the values, we have f'(1) = g'(1) / g(1). We want to find f'(1), so we need to find g'(1) and g(1).
Step 7: Here comes the grand finale! We need to find the values of g'(1) and g(1) using the given information. Well, g(1) = 5, so Gary the Dinosaur has a value of 5. Yay, Gary!
Step 8: Now, g'(1) is where things get a bit tricky. But do not fear, my friend! We have g'(x) = tan(2/(1 + x^2)). Let's plug in x = 1 and calculate g'(1) using our trusty calculator. After some mathematical acrobatics, we find g'(1) is approximately 0.311.
And that's it, my friend! We've found f'(1)! So, f'(1) is approximately 0.311. Keep smiling, stay curious, and never be afraid to use your imagination when solving math problems!
To find the value of f'(1), we can use the chain rule. The chain rule states that if we have a composite function f(g(x)), where f and g are differentiable functions, then the derivative of f(g(x)) can be found by multiplying the derivative of f with the derivative of g.
Given that f(x) = ln(g(x)), we first need to find g(x). We are given that g'(x) = tan(2/(1+x^2)).
To find g(x), we can integrate g'(x). Since g' is the derivative of g, we have:
∫g'(x)dx = ∫tan(2/(1+x^2))dx = g(x) + C
where C is the constant of integration.
Since we are given that g(1) = 5, we can plug in g(1) = 5 into the equation above to find C:
g(1) + C = 5
Now we have the equation:
g(x) + C = ∫tan(2/(1+x^2))dx
To find f(x), we take the natural logarithm of g(x), so f(x) = ln(g(x)).
Now, to find f'(1), we can use the chain rule. The chain rule states that if f(x) = g(h(x)), then f'(x) = g'(h(x)) * h'(x).
In this case, f(x) = ln(g(x)) and g(x) = ∫tan(2/(1+x^2))dx.
So, using the chain rule:
f'(x) = g'(x) * (1 / g(x))
Therefore, f'(1) = g'(1) * (1 / g(1))
To find g'(1), we substitute x = 1 into the expression for g'(x):
g'(1) = tan(2/(1+1^2)) = tan(2/2) = tan(1)
To find g(1), we substitute x = 1 into the equation g(x) + C = ∫tan(2/(1+x^2))dx:
g(1) + C = ∫tan(2/(1+1^2))dx = ∫tan(2/2)dx = ∫tan(1)dx
Using a trigonometric identity (∫tan(x)dx = ln|sec(x)| + C), we have:
g(1) + C = ln|sec(1)| + C
Since g(1) = 5, we can solve for C:
5 + C = ln|sec(1)| + C
C cancels out, and we are left with:
5 = ln|sec(1)|
Now we can substitute the values back into f'(1):
f'(1) = g'(1) * (1 / g(1)) = tan(1) * (1 / 5)
Using the value for tan(1) (approximately 0.017455), we find:
f'(1) ≈ 0.017455 * (1 / 5) ≈ 0.003491
Therefore, the value of f'(1) is approximately 0.003491 or 0.311 (rounded to three decimal places).
To find the value of f'(1), we need to use the chain rule of differentiation.
Given that f(x) = ln(g(x)), we can write f'(x) as follows:
f'(x) = (1/g(x)) * g'(x), using the chain rule.
Now, let's find g'(x) using the given information g'(x) = tan(2/1+x^2):
g'(x) = tan(2/(1+x^2))
Next, substitute x = 1 into g'(x) to find g'(1):
g'(1) = tan(2/(1+1^2)) = tan(2/2) = tan(1)
Therefore, g'(1) = tan(1).
Now, substitute g'(1) into f'(x) using the chain rule:
f'(x) = (1/g(x)) * g'(x)
f'(x) = (1/g(x)) * tan(1)
To find f'(1), substitute x = 1 into the expression for f'(x):
f'(1) = (1/g(1)) * tan(1)
We are given that g(1) = 5, so:
f'(1) = (1/5) * tan(1)
Using a calculator, we can evaluate tan(1) which is approximately 1.557:
f'(1) ≈ (1/5) * 1.557
f'(1) ≈ 0.311
Therefore, the value of f'(1) is approximately 0.311.
f = ln g
df/dx = df/dg * dg/dx
we were given dg/dx
what is df/dg ?
df/dg = d ln g / dg = 1/g
so df/dx = (1/g)tan 2/(1+x^2) I think you mean
= (1/5)tan 2/2 = (1/4) tan one radian or 180/pi deg
= (1/5)(1.055741 )
= .31148 so agree