Differentiate x5^sqrt(x).
Help please!
not clear
is it
x(5)^√x ??
Yes, basically, "x times 5 to the square root of x".
use the product rule,
you will have to know how to differentiate a function of the type y = a^u, where a is any positive number
so
y = (x)(5^√x)
dy/dx = x (5^√x)(ln5)(1/2)(x^)(-1/2) + 5^√x
Ah yes! Ok, I see what's going on. Thanks!
To differentiate the function \(f(x) = x^5 \sqrt{x}\), we can use the product rule and chain rule.
The product rule states that if we have two functions, \(u(x)\) and \(v(x)\), the derivative of their product is given by \( (u \cdot v)' = u' \cdot v + u \cdot v'\).
Let's apply the product rule to the function \(f(x) = x^5 \sqrt{x}\).
First, let's identify \(u(x) = x^5\) and \(v(x) = \sqrt{x}\).
Next, we need to find the derivatives of \(u(x)\) and \(v(x)\).
The derivative of \(u(x)\) with respect to \(x\) can be found by applying the power rule: \(u'(x) = 5x^{5-1} = 5x^4\).
The derivative of \(v(x)\) with respect to \(x\) can be found using the chain rule. Let's rewrite \(v(x)\) as \(x^{1/2}\), where the exponent is \(1/2\). Applying the chain rule, we have \(v'(x) = (1/2)x^{(1/2) - 1} = (1/2)x^{-1/2} = \frac{1}{2\sqrt{x}}\).
Now, we can apply the product rule:
\(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\)
\(f'(x) = 5x^4 \cdot \sqrt{x} + x^5 \cdot \frac{1}{2\sqrt{x}}\)
Simplifying,
\(f'(x) = 5x^4 \cdot \sqrt{x} + \frac{x^5}{2\sqrt{x}}\)
This is the derivative of the given function \(f(x) = x^5 \sqrt{x}\).