What is (\partial)/(\partial x)(b/(2b-k^2-1-2k*x)) ?

The answer to (\partial)/(\partial x)(b/(2b-k^2-1-2k*x)) is (2kb)/((2b-k^2-1-2kx)^2)

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(\partial)/(\partial x)(b/(2b-k^2-1-2k*x))

\frac{\partial}{\partial x} (\frac{b}{2 b - k ^{2} - 1 - 2 k \cdot x})

\frac{2 kb}{( 2 b - k ^{2} - 1 - 2 k x ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{b}{2 b - k ^{2} - 1 - 2 k x})

Treat

k, b

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= b \frac{\partial}{\partial x} (\frac{1}{2 b - k ^{2} - 1 - 2 k x})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= b \frac{\partial}{\partial x} ((2 b - k^{2} - 1 - 2 k x)^{- 1})

Apply the chain rule:

- \frac{1}{( 2 b - k ^{2} - 1 - 2 k x ) ^{2}} \frac{\partial}{\partial x} (2 b - k^{2} - 1 - 2 k x)

= - \frac{1}{( 2 b - k ^{2} - 1 - 2 k x ) ^{2}} \frac{\partial}{\partial x} (2 b - k^{2} - 1 - 2 k x)

\frac{\partial}{\partial x} (2 b - k^{2} - 1 - 2 k x) = - 2 k

= b (- \frac{1}{( 2 b - k ^{2} - 1 - 2 k x ) ^{2}} (- 2 k))

Simplify

b (- \frac{1}{( 2 b - k ^{2} - 1 - 2 k x ) ^{2}} (- 2 k)) : \frac{2 kb}{( 2 b - k ^{2} - 1 - 2 k x ) ^{2}}

= \frac{2 kb}{( 2 b - k ^{2} - 1 - 2 k x ) ^{2}}

\int 2 sin^{2} (x co) s^{2} x d x

7 sec (x) tan (x)

y^{^{'}} = \frac{[ x y ^{2} + ( y ) e ^{x} ]}{e ^{x}}

\frac{d}{d x} (\frac{x}{x ^{- 1} + 10})

f (x) = \frac{5 ^{x}}{x ^{5}}

What is (\partial)/(\partial x)(b/(2b-k^2-1-2k*x)) ?
The answer to (\partial)/(\partial x)(b/(2b-k^2-1-2k*x)) is (2kb)/((2b-k^2-1-2kx)^2)