prove that
d/dx 4x .√(x + √x) = 6x+5 (x)1/2/√(x + √x)
solution is
d/dx 4x.[x+(x)1/2]1/2 = d/dx 4x.[x+(x)^1/2]^1/2
d/dx 4x.[x+(x)^1/2]^1/2
Product Rule
= 4x[1/2(x+(x)^1/2)^-1/2 * (1+1/2x^-1/2) +[x+(x)^1/2]^1/2*4
d/dx(4 x sqrt(x+sqrt(x))) = (2 (1/(2 sqrt(x))+1) x)/sqrt(x+sqrt(x))+4 sqrt(x+sqrt(x))
am i write?
I cant follow your work. Try it here. http://cs.jsu.edu/~leathrum/Mathlets/derivcalc.html
Almost! There are a couple of minor mistakes in your answer. Here is the correct breakdown of the solution:
We are given the expression:
d/dx (4x√(x + √x))
To differentiate this expression, we can use the product rule, which states that if we have two functions u(x) and v(x) multiplied together, their derivative is given by:
d/dx (u(x)v(x)) = u'(x)v(x) + u(x)v'(x)
Let's apply the product rule to our expression:
u(x) = 4x
v(x) = √(x + √x)
First, let's find the derivatives of u(x) and v(x):
u'(x) = d/dx (4x) = 4
v'(x) = d/dx (√(x + √x))
To find the derivative of v(x), we can use the chain rule. The chain rule states that for a function f(g(x)), its derivative is given by:
d/dx (f(g(x))) = f'(g(x)) * g'(x)
Let's apply the chain rule to v(x):
f(x) = √x
g(x) = x + √x
f'(x) = d/dx (√x) = 1/(2√x)
g'(x) = d/dx (x + √x) = 1 + 1/(2√x)
Now, let's substitute these derivatives back into the product rule expression:
d/dx (4x√(x + √x)) = u'(x)v(x) + u(x)v'(x)
= 4 * √(x + √x) + 4x * [1/(2√x)] * [1 + 1/(2√x)]
= 4√(x + √x) + 2x/(√x) * [1 + 1/(2√x)]
= 4√(x + √x) + 2√x * [1 + 1/(2√x)]
= 4√(x + √x) + 2√x + 1
So, the correct derivative is:
d/dx (4x√(x + √x)) = 4√(x + √x) + 2√x + 1