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Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
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$\frac{\frac{d}{dk}\left(\left(k+1\right)^2\right)\left(2\left(k+1\right)-1\right)\left(2\left(k+1\right)+1\right)-\left(k+1\right)^2\frac{d}{dk}\left(\left(2\left(k+1\right)-1\right)\left(2\left(k+1\right)+1\right)\right)}{\left(\left(2\left(k+1\right)-1\right)\left(2\left(k+1\right)+1\right)\right)^2}$
Learn how to solve differential calculus problems step by step online. Find the derivative of ((k+1)^2)/((2(k+1)-1)(2(k+1)+1)). Apply the quotient rule for differentiation, which states that if f(x) and g(x) are functions and h(x) is the function defined by {\displaystyle h(x) = \frac{f(x)}{g(x)}}, where {g(x) \neq 0}, then {\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}. The power of a product is equal to the product of it's factors raised to the same power. Apply the product rule for differentiation: (f\cdot g)'=f'\cdot g+f\cdot g', where f=. Simplify the product -(\frac{d}{dk}\left(2\left(k+1\right)-1\right)\left(2\left(k+1\right)+1\right)+\left(2\left(k+1\right)-1\right)\frac{d}{dk}\left(2\left(k+1\right)+1\right)).