What Is the Second Derivative of f(x^3) Given Complex Function Relationships?

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The discussion focuses on finding the second derivative of f(x^3) given the relationships between f(x) and g(x). It starts with the equations df(x)/dx = g(x) and dg(x)/dx = f(x^2), leading to the first derivative of f(x) being equal to g(x). The chain rule is applied to express these derivatives in terms of x^3 and x^6, facilitating the calculation of the second derivative. The final expression for the second derivative incorporates both f(x^6) and g(x^3) along with their respective second derivatives. This complex differentiation problem emphasizes the necessity of a solid grasp of derivative properties and the chain rule.
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If \frac{\,df(x)}{\,dx} = g(x) and \frac{\,dg(x)}{\,dx} = f(x^2)
Then what is \frac{\,d^2f(x^3)}{\,dx^2}?
 
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differentiate \frac{\,df(x)}{\,dx} = g(x)
u will get \frac{\,d^2f(x)}{\,dx^2}=\frac{dg(x)}{dx}=f(x^2)

Is this enough
 


This is a challenging differentiation problem that requires a thorough understanding of the chain rule and the properties of derivatives. To solve this, we can break it down into smaller steps.

First, we can use the chain rule to rewrite the given equations as:

\frac{\,df(x)}{\,dx} = g(x) = g(x^3) \cdot \frac{\,dx^3}{\,dx}

and

\frac{\,dg(x)}{\,dx} = f(x^2) = f(x^6) \cdot \frac{\,dx^6}{\,dx}

Next, we can apply the chain rule again to find the second derivative:

\frac{\,d^2f(x^3)}{\,dx^2} = \frac{\,d}{\,dx} (g(x^3) \cdot \frac{\,dx^3}{\,dx}) = \frac{\,dg(x^3)}{\,dx} \cdot \frac{\,dx^3}{\,dx} + g(x^3) \cdot \frac{\,d^2x^3}{\,dx^2}

Similarly, we can find the second derivative of g(x^3):

\frac{\,d^2g(x^3)}{\,dx^2} = \frac{\,d}{\,dx} (f(x^6) \cdot \frac{\,dx^6}{\,dx}) = \frac{\,df(x^6)}{\,dx} \cdot \frac{\,dx^6}{\,dx} + f(x^6) \cdot \frac{\,d^2x^6}{\,dx^2}

Substituting these values into our original equation, we get:

\frac{\,d^2f(x^3)}{\,dx^2} = (f(x^6) \cdot \frac{\,d^2x^6}{\,dx^2}) \cdot \frac{\,dx^3}{\,dx} + (g(x^3) \cdot \frac{\,d^2x^3}{\,dx^2}) \cdot \frac{\,dx^6}{\,dx}

Since we know that \frac{\,d^2x^n}{\,dx^2} = n(n-1)x^{n-2}, we
 
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