If the function f is defined by the formula f(x)=x^3, then:
f ' (3x^2) = ?
I don't really get the question, but doesn't this basically mean the differential of the differential of f(x) as 3x^2 is the differential of x^3, eg 6x. But this doesn't seem right as its too easy almost.
Basic statistics::
shows that variances of independent (uncorrelated) variables are additive. . because of the awkwardness of the absolute value function in calculus.
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Mathematics Education::
Mar 5, 2005 Single Variable Calculus 2. Multivariate Calculus 3. . Function Theory of One Complex Variable by Robert E. Greene and Steven G. Krantz.
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Hope you can help someone! thanks
Tom
f(x)=x³
f ' (x)=3x²
f ' (3x²)=3(3x²)²
=3(9x^4)
=27x^4
It's telling you to find f prime of x which is f'(x)=3x² then substitute in 3x² for x so you get f'(3x²)=3(3x²)² which simplifies to the answer given above.
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