How do you differentiate # 3/4 * (2x^3 + 3x)^(-1/4)#?

Question

Charles Arocha · Accepted Answer

#(-18x^2+9)/(16(2x^3+3x)^(5/4)# We can factor a constant out of the equation: #3/4 * [(2x^3+3x)^(-1/4)]# Then we apply the chain rule: #d/dx[f(g(x))] -> f'(g(x)) * g'(x)# #d/dx[(2x^3+3x)^(-1/4)] -> -1/4(2x^3+3x)^(-5/4) # #d/dx[2x^3+3x] -> 6x^2+3# We multiply the outside derivative and inside derivative together: #3/4 * [-1/4(2x^3+3x)^(-5/4) * (6x^2+3)]# Bring the negative exponent down below: #3/4*-(6x^2+3)/(4(2x^3+3x)^(5/4)# Multiply the constant across: #(-18x^2+9)/(16(2x^3+3x)^(5/4)#

William Abdalla · Answer

undefined To differentiate the expression $ \frac{3}{4} \cdot (2x^3 + 3x)^{-\frac{1}{4}} $, you can use the chain rule. The derivative can be calculated as follows:

$$ \frac{d}{dx}\left(\frac{3}{4} \cdot (2x^3 + 3x)^{-\frac{1}{4}}\right) = \frac{3}{4} \cdot \frac{d}{dx}\left((2x^3 + 3x)^{-\frac{1}{4}}\right) $$

To find the derivative of $ (2x^3 + 3x)^{-\frac{1}{4}} $, we use the chain rule:

$$ \frac{d}{dx}\left((2x^3 + 3x)^{-\frac{1}{4}}\right) = -\frac{1}{4}(2x^3 + 3x)^{-\frac{1}{4} - 1} \cdot \frac{d}{dx}(2x^3 + 3x) $$

$$ = -\frac{1}{4}(2x^3 + 3x)^{-\frac{5}{4}} \cdot (6x^2 + 3) $$

Now, multiply this result by $ \frac{3}{4} $:

$$ \frac{3}{4} \cdot -\frac{1}{4}(2x^3 + 3x)^{-\frac{5}{4}} \cdot (6x^2 + 3) $$

$$ = -\frac{3}{16}(2x^3 + 3x)^{-\frac{5}{4}} \cdot (6x^2 + 3) $$

So, the derivative of $ \frac{3}{4} \cdot (2x^3 + 3x)^{-\frac{1}{4}} $ is $ -\frac{3}{16}(2x^3 + 3x)^{-\frac{5}{4}} \cdot (6x^2 + 3) $.