(d^(k)y)/(dx^(k))=(-1)^(k-1)((k-1)!a^(k))/((ax+1)^(n))

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Identify Function & Operation: Identify the function and the operation to be performed. We are asked to find the k-th derivative of the function y with respect to x, where y is given by the formula y = (-1)^(k-1)((k-1)!a^(k))/((ax+1)^(n)). This is a differentiation problem involving higher-order derivatives.Apply k-th Derivative: Apply the k-th derivative operation. To find the k-th derivative, we will use the general Leibniz rule for differentiation, which is an extension of the product rule to higher derivatives. However, since the function is already expressed in terms of k, we need to verify if the expression given is indeed the k-th derivative or if we need to perform additional differentiation steps.Analyze Given Expression: Analyze the given expression. The expression (-1)^(k-1)((k-1)!a^(k))/((ax+1)^(n)) seems to be the result of applying the k-th derivative operation to some original function. The presence of the factorial (k-1)! and the power of a suggest that this is the result of differentiating a function involving (ax+1) to some power. The negative sign raised to the power of (k-1) indicates alternating signs for successive derivatives.Determine Final Answer: Determine if the expression is the final answer. Since the problem statement does not provide the original function but directly gives us what appears to be the k-th derivative, we do not need to perform any further differentiation. The expression given is the final form of the k-th derivative.Verify for Errors: Verify the expression for any possible common differentiation errors. Common errors in differentiation include incorrect application of the chain rule, product rule, or power rule. In this case, since we are not explicitly performing the differentiation, we need to ensure that the expression follows the pattern of derivatives for a function of the form (ax+1)^(-n). The pattern of alternating signs and the factorial term suggest that the differentiation has been correctly applied.Conclude Solution: Conclude the solution. The given expression is already in the form of the k-th derivative, so no further differentiation is required. The solution to the problem is the expression itself.

$\left(\frac{d^{k}y}{dx^{k}}\right) = (-1)^{k-1}\frac{(k-1)!a^{k}}{(ax+1)^{n}}$

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(dkydxk)=(−1)k−1(k−1)!ak(ax+1)n\left(\frac{d^{k}y}{dx^{k}}\right) = (-1)^{k-1}\frac{(k-1)!a^{k}}{(ax+1)^{n}}(dxkdky​)=(−1)k−1(ax+1)n(k−1)!ak​

$\left(\frac{d^{k}y}{dx^{k}}\right) = (-1)^{k-1}\frac{(k-1)!a^{k}}{(ax+1)^{n}}$