What is the derivative of 2/((x-3t^2+1)) ?

The derivative of 2/((x-3t^2+1)) is -(4(x-3t))/(((x-3t)^2+1)^2)

What is the first derivative of 2/((x-3t^2+1)) ?

The first derivative of 2/((x-3t^2+1)) is -(4(x-3t))/(((x-3t)^2+1)^2)

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\frac{d}{d x} (\frac{2}{( x - 3 t ) ^{2} + 1})

- \frac{4 ( x - 3 t )}{( ( x - 3 t ) ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2}{( x - 3 t ) ^{2} + 1})

Treat

t

as a constant

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{1}{( x - 3 t ) ^{2} + 1})

Apply exponent rule:

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

= 2 \frac{d}{d x} (((x - 3 t)^{2} + 1)^{- 1})

Apply the chain rule:

- \frac{1}{( ( x - 3 t ) ^{2} + 1 ) ^{2}} \frac{d}{d x} ((x - 3 t)^{2} + 1)

= - \frac{1}{( ( x - 3 t ) ^{2} + 1 ) ^{2}} \frac{d}{d x} ((x - 3 t)^{2} + 1)

\frac{d}{d x} ((x - 3 t)^{2} + 1) = 2 (x - 3 t)

= 2 - \frac{1}{( ( x - 3 t ) ^{2} + 1 ) ^{2}} \cdot 2 (x - 3 t)

Simplify

2 - \frac{1}{( ( x - 3 t ) ^{2} + 1 ) ^{2}} \cdot 2 (x - 3 t) : - \frac{4 ( x - 3 t )}{( ( x - 3 t ) ^{2} + 1 ) ^{2}}

= - \frac{4 ( x - 3 t )}{( ( x - 3 t ) ^{2} + 1 ) ^{2}}

x \to 1 lim (x (x - 2))

x \to 0 + lim (x)

n = 0 \sum \infty sin (\frac{nπ}{7})

\frac{d}{d x} (sinh (9 x))

\int_{0}^{1} (e^{x} + e^{- x})^{2} d x

What is the derivative of 2/((x-3t^2+1)) ?
The derivative of 2/((x-3t^2+1)) is -(4(x-3t))/(((x-3t)^2+1)^2)
What is the first derivative of 2/((x-3t^2+1)) ?
The first derivative of 2/((x-3t^2+1)) is -(4(x-3t))/(((x-3t)^2+1)^2)