Differentiate x cosh x - sinh x with respect to x
Understand the Problem
The question is asking us to find the derivative of the expression 'x cosh x - sinh x' with respect to the variable 'x'. This involves applying the product and chain rule of differentiation.
Answer
The derivative is $x \sinh x$.
Answer for screen readers
The derivative of the expression $x \cosh x - \sinh x$ is $x \sinh x$.
Steps to Solve
- Differentiate the first term using the product rule
The expression is $x \cosh x$. To differentiate this, we apply the product rule, which states that if you have two functions $u(x)$ and $v(x)$, then the derivative is given by:
$$ (uv)' = u'v + uv' $$
Here, let $u = x$ and $v = \cosh x$. Then, we differentiate:
- $u' = 1$
- $v' = \sinh x$
Now we can apply the product rule:
$$ (x \cosh x)' = u'v + uv' = 1 \cdot \cosh x + x \cdot \sinh x = \cosh x + x \sinh x $$
- Differentiate the second term
The second term is $-\sinh x$. Its derivative with respect to $x$ is straightforward:
$$ (-\sinh x)' = -\cosh x $$
- Combine the derivatives
Now we combine the derivatives of both terms:
$$ \frac{d}{dx}(x \cosh x - \sinh x) = \cosh x + x \sinh x - \cosh x $$
Notice that the $\cosh x$ terms cancel each other out.
- Simplify the expression
The remaining expression is:
$$ x \sinh x $$
The derivative of the expression $x \cosh x - \sinh x$ is $x \sinh x$.
More Information
In calculus, the hyperbolic functions such as $\cosh x$ and $\sinh x$ are analogous to the circular functions. The product and chain rules are essential for finding derivatives of products of functions, especially when dealing with hyperbolic functions.
Tips
- Forgetting to apply the product rule correctly when differentiating $x \cosh x$.
- Neglecting the negative sign in front of $\sinh x$.
- Overlooking the cancellation of terms when combining derivatives.
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