Given that frac{d}{dx}f(x) = f'(x). The relat | vedclass

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(B) Given the relationship $f'(a + b) = f'(a) + f'(b)$.Let us test the options:For option $(A)$,$f(x) = x$,then $f'(x) = 1$. Thus,$f'(a+b) = 1$ and $f

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$x^2$

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(B) Given the relationship $f'(a + b) = f'(a) + f'(b)$.Let us test the options:For option $(A)$,$f(x) = x$,then $f'(x) = 1$. Thus,$f'(a+b) = 1$ and $f'(a) + f'(b) = 1 + 1 = 2$. Since $1 
eq 2$,this is incorrect.For option $(B)$,$f(x) = x^2$,then $f'(x) = 2x$. Thus,$f'(a+b) = 2(a+b) = 2a + 2b$ and $f'(a) + f'(b) = 2a + 2b$. Since $2a + 2b = 2a + 2b$,this is correct.For option $(C)$,$f(x) = x^3$,then $f'(x) = 3x^2$. Thus,$f'(a+b) = 3(a+b)^2 = 3(a^2 + 2ab + b^2)$ and $f'(a) + f'(b) = 3a^2 + 3b^2$. These are not equal.Therefore,the correct function is $f(x) = x^2$.

Given that $\frac{d}{dx}f(x) = f'(x)$. The relationship $f'(a + b) = f'(a) + f'(b)$ is valid if $f(x)$ is equal to

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