Differentiate 10^cosh(√x) with respect to x.
Understand the Problem
The question is asking how to differentiate the function 10 raised to the hyperbolic cosine of the square root of x with respect to x. We will use the chain rule and properties of exponential functions to find the derivative.
Answer
$$ \frac{dy}{dx} = 10^{\cosh(\sqrt{x})} \cdot \ln(10) \cdot \frac{\sinh(\sqrt{x})}{2\sqrt{x}} $$
Answer for screen readers
The derivative of the function $y = 10^{\cosh(\sqrt{x})}$ with respect to $x$ is:
$$ \frac{dy}{dx} = 10^{\cosh(\sqrt{x})} \cdot \ln(10) \cdot \frac{\sinh(\sqrt{x})}{2\sqrt{x}} $$
Steps to Solve
- Identify the function to differentiate
The function we want to differentiate is given by:
$$ y = 10^{\cosh(\sqrt{x})} $$
- Apply the chain rule
To differentiate this function, we will use the chain rule. The chain rule states that if we have a composite function $f(g(x))$, then the derivative is given by $f'(g(x)) \cdot g'(x)$. Here, we can consider $f(u) = 10^u$ where $u = \cosh(\sqrt{x})$.
- Differentiate the outer function
The derivative of the outer function $f(u) = 10^u$ is:
$$ f'(u) = 10^u \ln(10) $$
So, applying this to our function, we have:
$$ f'(\cosh(\sqrt{x})) = 10^{\cosh(\sqrt{x})} \ln(10) $$
- Differentiate the inner function
Next, we need to find the derivative of the inner function $u = \cosh(\sqrt{x})$. We use the chain rule again:
- The derivative of $\cosh(v)$ is $\sinh(v)$, where $v = \sqrt{x}$.
- The derivative of $\sqrt{x}$ with respect to $x$ is $\frac{1}{2\sqrt{x}}$.
So, using the chain rule, we have:
$$ \frac{du}{dx} = \sinh(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} $$
- Combine the derivatives
Now we can combine our findings:
$$ \frac{dy}{dx} = f'(\cosh(\sqrt{x})) \cdot \frac{du}{dx} $$
Substituting our derivatives, we get:
$$ \frac{dy}{dx} = 10^{\cosh(\sqrt{x})} \ln(10) \cdot \left( \sinh(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \right) $$
- Write the final derivative
The final result is:
$$ \frac{dy}{dx} = 10^{\cosh(\sqrt{x})} \cdot \ln(10) \cdot \frac{\sinh(\sqrt{x})}{2\sqrt{x}} $$
The derivative of the function $y = 10^{\cosh(\sqrt{x})}$ with respect to $x$ is:
$$ \frac{dy}{dx} = 10^{\cosh(\sqrt{x})} \cdot \ln(10) \cdot \frac{\sinh(\sqrt{x})}{2\sqrt{x}} $$
More Information
This derivative shows how the function grows with respect to $x$. The presence of hyperbolic functions indicates that the behavior of the function could have interesting characteristics over different intervals of $x$.
Tips
- A common mistake is to forget to apply the chain rule correctly when differentiating composite functions.
- Another mistake is to miscalculate the derivatives of hyperbolic functions or the derivative of functions like $\sqrt{x}$.
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