If d/dx(f(x))=g(x) and d/dx(g(x))=f(x^2) , then d^2/dx^2(f(x^3))
, then d2dx2(f(x3))=
A f(x^6)
B g(x^3)
C 3x^2g(x^3)
D 9x^4f(x^6) + 6xg(x^3)
E f(x^6) + g(x^3)
I understood what the first half of the question said, but then I was confused by what the second half says. I'm confused on how to proceed with this question :(
(f’(x)=d/dx(f(x)), I use both forms while solving.)
d/dx(f(x))=g(x) is defined,
so d^2/d^2x(f(x))=g’(x)dx (chain rule).
d/dx(g(x))=f(x^2) is defined.
Now, we look at what we need to find, the second derivative of f(x^3)
d^2/d^2x f(x^3) = d/dx(f’(x^3)). Plug in using what is defined.
=d/dx(g(x^3)3x^2). Now use product rule to derivate.
= g(x^3)6x + g’(x^3)3x^2. Plug in g’(x)=f(x^2)
=g(x^3)6x+f((x^3)^2)(3x^2)(3x^2)
=g(x^3)6x+f(x^6)9x^4
With a little rearranging, we can see the answer is D. It’s important to not forget the chain rule in any step in this question, and to apply the product rule when we had to :)
To find the second derivative of \(f(x^3)\), we can use the chain rule of differentiation.
Let's start by finding the first derivative of \(f(x^3)\):
\(\frac{d}{dx}(f(x^3))\)
Now, substitute \(x^3\) as a new variable, let \(u = x^3\):
\(\frac{d}{dx}(f(u))\)
Since \(u = x^3\), we can express the derivative with respect to \(u\):
\(\frac{d}{du}(f(u))\cdot\frac{du}{dx}\)
The factor \(\frac{du}{dx}\) represents the derivative of \(u\) with respect to \(x\), which is \(3x^2\).
Thus, the first derivative becomes:
\(\frac{d}{dx}(f(x^3)) = \frac{d}{du}(f(u)) \cdot \frac{du}{dx} = \frac{d}{du}(f(u)) \cdot 3x^2\)
Given that \(\frac{d}{dx}(f(x)) = g(x)\), we can rewrite the expression as:
\(g(u) \cdot 3x^2\)
Now, we can proceed to finding the second derivative of \(f(x^3)\) by differentiating the expression \(\frac{d}{dx}(f(x^3))\) with respect to \(x\):
\(\frac{d}{dx}\left(g(u) \cdot 3x^2\right)\)
Using the product rule, we differentiate each term separately:
\(\frac{d}{du}(g(u)) \cdot \frac{du}{dx} \cdot 3x^2 + g(u) \cdot \frac{d}{dx}\left(3x^2\right)\)
The first term, \(\frac{d}{du}(g(u)) \cdot \frac{du}{dx} = \frac{d}{du}(g(u)) \cdot 3x^2\), is equal to \(\frac{d}{dx}(g(x^3)) \cdot 3x^2\).
The second term, \(\frac{d}{dx}\left(3x^2\right)\), is equal to \(6x\).
Hence, the second derivative, \(\frac{d^2}{dx^2}(f(x^3))\), can be calculated as:
\(\frac{d^2}{dx^2}(f(x^3)) = \frac{d}{dx}\left(g(x^3)\right) \cdot 3x^2 + 6x\)
Given that \(\frac{d}{dx}(g(x)) = f(x^2)\), we can substitute that expression into the equation:
\(\frac{d^2}{dx^2}(f(x^3)) = f\left({\left(x^3\right)}^2\right) \cdot 3x^2 + 6x\)
Simplifying further, \({\left(x^3\right)}^2\) becomes \(x^6\):
\(\frac{d^2}{dx^2}(f(x^3)) = f(x^6) \cdot 3x^2 + 6x\)
Therefore, the answer is \textbf{(E) f(x^6) + g(x^3)}.
Well, let's break it down step by step. We know that d/dx(f(x)) = g(x), and d/dx(g(x)) = f(x^2).
Now, to find d^2/dx^2(f(x^3)), we need to take the derivative of d/dx(f(x^3)). To do this, we can use the chain rule. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).
In our case, g(x) is f(x^3). Let's call it h(x) for now, just to make it clearer. So, g(x) = h(x) = f(x^3). We can rewrite the expression we're looking for as d^2/dx^2(h(x)).
Using the chain rule, d/dx(h(x)) = f'(x^3) * (x^3)' = 3x^2 * f'(x^3).
Now, we need to differentiate again. d^2/dx^2(h(x)) = d/dx(3x^2 * f'(x^3)).
Again, we can use the chain rule, but this time we have a product of two functions: 3x^2 and f'(x^3). Let's call 3x^2 "m(x)" and f'(x^3) "n(x)". So, we have m(x) * n(x).
Differentiating the product, we get d/dx(m(x) * n(x)) = m'(x) * n(x) + m(x) * n'(x).
In our case, m(x) = 3x^2 and n(x) = f'(x^3). Differentiating m(x) gives us m'(x) = 6x, and differentiating n(x) gives us n'(x) = f''(x^3) * (x^3)' = 3x^2 * f''(x^3).
Plugging everything back in, we get d^2/dx^2(h(x)) = 6x * f'(x^3) + 3x^2 * f''(x^3).
But remember, h(x) is f(x^3), so we can replace it back in to get d^2/dx^2(f(x^3)) = 6x * f'(x^3) + 3x^2 * f''(x^3).
So, the answer is C) 6x * f'(x^3) + 3x^2 * f''(x^3).
d/dx f(x^3) = df/dx * 3x^2
so,
d^2/dx^2 f(x^3) = d/dx (d/dx f(x^3))
= d/dx (df/dx * 3x^2)
= d/dx (df/dx)) * 3x^2 + 6x df/dx
It's beginning to look like D, right?