Why is $n = k+1$ in $\frac{d^ny}{dx^n} = (\frac{d}{dx})(\frac{d^{k-1}y}{dx^{k-1}})$ and not $\frac{d^{k+1}y}{dx^{k+1}}$?

Asked Nov 21 '20 at 03:46

Active Nov 21 '20 at 06:56

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The question stems from the following mathematical induction:

However, the solution I found here https://www.chegg.com/tutors/Calculus-questions/Use-the-principle-of-mathematical-induction-to-prove-that-if-y-1-1-x-then-dny-dxn-n-1-xn-+-1-for-n-Z+--J7DSJ/ states that $n = k+1$ is being used, yet a $\frac{d^{k-1}y}{dx^{k-1}}$ appears in the working out. Please help me understand

When I tired it myself anyway, I don't know how to get $\frac{d^{k+1}y}{dx^{k+1}}$ back in the form $\frac{d^ky}{dx^k}$ anyway (to substitute the assumption when $n=k$.

asked Nov 21 '20 at 03:46

user71207

1,543

2

$\frac {d^{k+1}y}{dx^{k+1}} = \frac d{dx}\left(\frac {d^k y}{dx^k}\right)$ – player3236 Nov 21 '20 at 03:53
But there is an extra d in the denominator on the RHS! – user71207 Nov 21 '20 at 03:55
These are not fractions; they are operators. – player3236 Nov 21 '20 at 03:56
$$\frac{d(\frac{dy}{dx})}{dx}=\frac{d^2y}{dx^2} \ne \frac{d^2y}{d^2y^2}$$ – DatBoi Nov 21 '20 at 03:59
1

Same as in player3236 comment with slight difference $$\frac {d^{k+1}y}{dx^{k+1}} = \frac {d\left(\frac {d^k y}{dx^k}\right)}{dx}$$ – zkutch Nov 21 '20 at 04:05

Why is $n = k+1$ in $\frac{d^ny}{dx^n} = (\frac{d}{dx})(\frac{d^{k-1}y}{dx^{k-1}})$ and not $\frac{d^{k+1}y}{dx^{k+1}}$?

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