find the derivative of the function y=(3x+4)^5(4x+1)^-2
3x+4*5/4x+1*2
4x+1*2d/dx(3x+4)*5....................................... matlb quotientrule apply krlo ok janu i am itself 3rd year student
online, we generally use ^ for powers, and * for multiplication. Why bother changing the way it was posted?
And the quotient rule says
5(3x+4)^4(3)(4x+1)^2 - (3x+4)^5(2(4x+1)(4))
-----------------------------------------
(4x+1)^4
(3x+4)^4 (15(4x+1)-8(3x+4))
------------------------
(4x+1)^3
(36x-17)(3x+4)^4 / (4x+1)^3
To find the derivative of the function y = (3x+4)^5(4x+1)^-2, we can use the product rule and the chain rule.
The product rule states that if we have two functions u(x) and v(x), the derivative of their product u(x)*v(x) with respect to x is given by:
(d/dx) [u(x) * v(x)] = u'(x) * v(x) + u(x) * v'(x)
Let's apply the product rule to the given function.
First, let's determine u(x) and v(x):
u(x) = (3x+4)^5
v(x) = (4x+1)^-2
Next, let's find the derivatives u'(x) and v'(x) using the chain rule:
u'(x) = 5(3x+4)^4 * 3
v'(x) = -2(4x+1)^-3 * 4
Now, we can substitute these values into the product rule formula:
(d/dx) [y] = u'(x) * v(x) + u(x) * v'(x)
= [5(3x+4)^4 * 3] * [v(x)] + [u(x)] * [-2(4x+1)^-3 * 4]
Simplifying further:
(d/dx) [y] = 15(3x+4)^4 * [v(x)] - 8(4x+1)^-3 * [u(x)]
Substituting u(x) = (3x+4)^5 and v(x) = (4x+1)^-2:
(d/dx) [y] = 15(3x+4)^4 * (4x+1)^-2 - 8(4x+1)^-3 * (3x+4)^5
Therefore, the derivative of y = (3x+4)^5(4x+1)^-2 is given by:
(d/dx) [y] = 15(3x+4)^4 * (4x+1)^-2 - 8(4x+1)^-3 * (3x+4)^5