Differentiate e^(ax) sinh(bx) with respect to x
Understand the Problem
The question is asking to find the derivative of the function e^(ax) * sinh(bx) with respect to x. This involves applying the product rule and the derivatives of exponential and hyperbolic sine functions.
Answer
The derivative is given by: $$ \frac{dy}{dx} = e^{ax}(a \sinh(bx) + b \cosh(bx)) $$
Answer for screen readers
The derivative of the function $e^{ax} \cdot \sinh(bx)$ with respect to $x$ is:
$$ \frac{dy}{dx} = e^{ax}(a \sinh(bx) + b \cosh(bx)) $$
Steps to Solve
- Identify the functions involved
The function we are working with is $y = e^{ax} \cdot \sinh(bx)$. Here, we can see that we have two functions: $u = e^{ax}$ and $v = \sinh(bx)$.
- Apply the Product Rule
To find the derivative of a product of two functions, we use the product rule, which states that:
$$ \frac{dy}{dx} = u'v + uv' $$
- Differentiate the first function
Now, we differentiate $u = e^{ax}$:
$$ u' = \frac{d}{dx}(e^{ax}) = ae^{ax} $$
- Differentiate the second function
Next, we differentiate $v = \sinh(bx)$:
$$ v' = \frac{d}{dx}(\sinh(bx)) = b \cosh(bx) $$
- Combine the results using the Product Rule
Now, substituting our derivatives back into the product rule formula:
$$ \frac{dy}{dx} = u'v + uv' = ae^{ax} \cdot \sinh(bx) + e^{ax} \cdot b \cosh(bx) $$
- Factor the expression
Finally, we can factor $e^{ax}$ out from the expression:
$$ \frac{dy}{dx} = e^{ax}(a \sinh(bx) + b \cosh(bx)) $$
The derivative of the function $e^{ax} \cdot \sinh(bx)$ with respect to $x$ is:
$$ \frac{dy}{dx} = e^{ax}(a \sinh(bx) + b \cosh(bx)) $$
More Information
The derivative involves applying the product rule and requires knowledge of differentiation rules for exponential and hyperbolic functions. The hyperbolic sine function, $\sinh$, is related to exponential functions and often appears in problem-solving contexts in calculus.
Tips
- Forgetting to apply the product rule correctly can lead to an incorrect derivative.
- Not differentiating each part of the product separately may cause errors.
- Miscomputing the derivatives of $\sinh(bx)$ or $e^{ax}$.
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