given that f, g, and h are differentiable functions f(g(h(x))) = x what is h'(x) in terms of f, g, f', g', and h'?
I plugged in equations on my own like
h(x)= x^4
g(x) = 4th root of x
f(x)= x^2
i need more assistance on how to proceed thnx
To find h'(x) in terms of f, g, f', g', and h', we can use the chain rule.
The chain rule states that if we have a composite function, such as f(g(h(x))), the derivative of the composite function is the product of the derivatives of each individual function multiplied together. In other words:
(d/dx) f(g(h(x))) = f'(g(h(x))) * g'(h(x)) * h'(x)
Let's apply the chain rule to our problem:
Given:
f(x) = x^2
g(x) = (fourth root of x)
h(x) = x^4
We are interested in finding h'(x).
Step 1: Find g(h(x)) and f(g(h(x))).
g(h(x)) = g(x^4) = (fourth root of x^4) = x
f(g(h(x))) = f(x) = x^2
Step 2: Find the derivatives of f(x), g(x), and h(x).
f'(x) = 2x (Power rule for differentiation)
g'(x) = 1/(4 * x^(3/4)) (Power rule and chain rule for differentiation)
h'(x) = 4x^(3) (Power rule for differentiation)
Step 3: Apply the chain rule.
(d/dx) f(g(h(x))) = f'(g(h(x))) * g'(h(x)) * h'(x)
We know that f(g(h(x))) = x^2. So,
2x^2 = f'(g(h(x))) * g'(h(x)) * h'(x)
Since f'(x) = 2x, g'(x) = 1/(4 * x^(3/4)), and h'(x) = 4x^(3), substituting these values into the equation:
2x^2 = (2x) * (1/(4 * x^(3/4))) * (4x^(3))
Simplifying,
2x^2 = 2x^(13/4)
Now, to solve for h'(x), we divide both sides of the equation by 2x^2:
h'(x) = (2x^(13/4))/(2x^2)
Simplifying further,
h'(x) = x^(13/4 - 2)
h'(x) = x^(5/4)
Thus, in terms of f, g, f', g', and h', h'(x) is equal to x^(5/4).