Let $$ y=x^{x}+\cos (x) $$. Find $$ y^{\prime} $$.

Question

UpStudy AI · Accepted Answer

To find the derivative $$ y' $$ of the function $$ y = x^x + \cos(x) $$, we will differentiate each term separately.

1. **Differentiate $$ x^x $$**:
   We can use logarithmic differentiation for $$ x^x $$. Let $$ z = x^x $$. Taking the natural logarithm of both sides, we have:
   $$
   \ln z = x \ln x
   $$
   Now, differentiate both sides with respect to $$ x $$:
   $$
   \frac{1}{z} \frac{dz}{dx} = \ln x + 1
   $$
   Therefore, we can express $$ \frac{dz}{dx} $$ as:
   $$
   \frac{dz}{dx} = z(\ln x + 1) = x^x (\ln x + 1)
   $$
   Thus, the derivative of $$ x^x $$ is:
   $$
   \frac{d}{dx}(x^x) = x^x (\ln x + 1)
   $$

2. **Differentiate $$ \cos(x) $$**:
   The derivative of $$ \cos(x) $$ is:
   $$
   \frac{d}{dx}(\cos x) = -\sin x
   $$

3. **Combine the derivatives**:
   Now we can combine the derivatives of both terms to find $$ y' $$:
   $$
   y' = \frac{d}{dx}(x^x) + \frac{d}{dx}(\cos x) = x^x (\ln x + 1) - \sin x
   $$

Thus, the derivative $$ y' $$ is:
$$
\boxed{x^x (\ln x + 1) - \sin x}
$$
The derivative $$ y' $$ is $$ x^x (\ln x + 1) - \sin x $$.

Let . Find .

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions