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f'(x) = -1/x^{5}

The equation above can be written as:

f'(x) = -x^{-5}

Integrate both sides with respect to x

∫f'(x) = ∫-x^{-5}

∫f'(x) = -∫x^{-5}

f(x) = -(x^{-5 + 1}) / -5+1 + c

where c is the integration constant.

f(x) = -x^{-4}/-4 + c

f(x) = x^{-4}/4 + c [The negative sign in the numerator and denominator cancels each other out].

f(x) = 1/4x^{4} + c ----------- eqn(1)

At f(-1) = -7/4 [that is, substitute x = -1 and equate to -7/4]

eqn(1) becomes,

-7/4 = 1/4(-1)^{4} + c

-7/4 = 1/4 + c

subtract 1/4 from both sides.

c = -7/4 - 1/4

c = -2

Thus, eqn(1) becomes:

f(x) = 1/4x^{4} - 2, x < 0

**Ans D**

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