If `x=f(t)` and `y=g(t)` , then `(d^2y)/(dx^2)` is equal to `(f^(prime)g"-g'\ f")/((f^(prime))^3)` (b) `(f^(prime)g"-g'\ f")/((f^(prime))^2)` (c) `(g")/(f")` (d) `(f^g'-g"\ f')/((g^(prime))^3)`

Explanation

`x=f(t), y=g(t)`
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frac{d x}{d t}=f^{prime}(t) `
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frac{d y}{d t}=g^{prime}(t) `
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frac{d y}{d x}=frac{frac{d y}{d t}}{frac{d x}{d t}}=frac{g^{prime}(t)}{f^{prime}(t)} `
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frac{d^{2} y}{d x^{2}}=frac{d}{d x}(frac{g^{prime}(t)}{f^{prime}(t)}) `
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=frac{f^{prime}(t) cdot g^{prime prime}(t)-g g^{prime}(t) cdot f^{prime prime}(t)}{(f^{prime}(t))^{2}} cdot frac{d t}{d x}`
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=frac{f^{prime}(t) cdot g^{prime prime}(t)-g^{prime}(t) cdot f^{prime prime}(t)}{(f^{prime}(t))^{2}} cdot frac{1}{f^{prime}(t)} `
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=frac{f^{prime}(t) cdot g^{prime prime}(t)-g^{prime}(t) cdot f^{prime prime}(t)}{[f^{prime}(t)]^{3}}`
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