How do you calculate the derivative of #(4x- 3)/(sqrt(2x^2 +1))#?

Question

Ezekiel Alexander · Accepted Answer

You can use the product rule to find the derivative.

Rewrite the function as follows

#y=(4x-3)(2x^2+1)^(-1/2) #

Using the product rule

#dy/dx=4(2x^2+1)^(-1/2)+(4x-3)(-1/2)(2x^2+1)^(-3/2)(4x)#

#dy/dx=4(2x^2+1)^(-1/2)-2x(4x-3)(2x^2+1)^(-3/2) #

#dy/dx=(2x^2+1)^(-3/2)[4(2x^2+1)-2x(4x-3)]#

#dy/dx=(2x^2+1)^(-3/2)[8x^2+4-8x^2+6x] #

#dy/dx=(2x^2+1)^(-3/2)[4+6x] #

#dy/dx=2(2x^2+1)^(-3/2)[2+3x] #

#dy/dx=(2(2+3x))/(2x^2+1)^(3/2) #

Aurora Alvarado · Answer

undefined To calculate the derivative of $(4x - 3)/(\sqrt{2x^2 + 1})$:

1. Apply the quotient rule:

$f'(x) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{(g(x))^2}$, where $f(x) = 4x - 3$ and $g(x) = \sqrt{2x^2 + 1}$.

2. Find the derivatives of $f(x)$ and $g(x)$:

$f'(x) = 4$ and $g'(x) = \frac{4x}{2\sqrt{2x^2 + 1}}$.

3. Substitute into the quotient rule formula:

$f'(x) = \frac{(2x\sqrt{2x^2 + 1}) - (4x - 3) \cdot \frac{4x}{2\sqrt{2x^2 + 1}}}{(2x^2 + 1)}$.

4. Simplify the expression:

$f'(x) = \frac{2x\sqrt{2x^2 + 1} - \frac{(4x)^2}{2\sqrt{2x^2 + 1}} + 3 \cdot \frac{4x}{2\sqrt{2x^2 + 1}}}{2x^2 + 1}$.

5. Combine like terms and simplify further if possible.