MHTCET 17 Apr 2026 Shift 2 Paper

Concept:Differentiating a reciprocal linear function $n$ times produces a result combining a factorial, a power of the inner coefficient, and an alternating sign.Explanation:First, express the function in power form:$y = \frac{1}{3x+5} = (3x+5)^{-1}$For any function $(ax+b)^{-1}$, the general formula for its $n$-th derivative is:$\frac{d^n}{dx^n}(ax+b)^{-1} = (-1)^n \cdot n! \cdot a^n \cdot (ax+b)^{-(n+1)}$This formula comes from applying the power rule and the chain rule repeatedly for each differentiation.In this problem, we have $a = 3$ and $b = 5$.We need the 9th derivative, so substitute $n = 9$:$\frac{d^9 y}{dx^9} = (-1)^9 \cdot 9! \cdot 3^9 \cdot (3x+5)^{-(9+1)}$Since $9+1 = 10$ and $(-1)^9 = -1$, the expression becomes:$\frac{d^9 y}{dx^9} = \frac{(-1)^9 \times 9! \times 3^9}{(3x+5)^{10}}$Comparing with the given options, this matches option C exactly.Answer:Option C: $\frac{(-1)^{9} \times 9!\times 3^{9}}{(3 x+5)^{10}}$