Question

Given c = sin(a - b ),$\mathit{b}\mathbf{=}\mathit{a}{\mathit{e}}^{\mathbf{2}\mathbf{a}}$, find $\frac{\mathbf{d}\mathbf{c}}{\mathbf{d}\mathbf{b}}$.

Short answer

The$\frac{dc}{db}$ is$\cos \left(a-b\right)\left(1-2b-e^{2a}\right)$ .

Step-by-step solution

Explanation of solution

The provided expressions are $c=\sin \left(a-b\right),b=a{e}^{2a}$.

Chain Rule

The number of functions in the composition affects how many differentiation steps are required when using the chain rule to get the derivative of composite functions.

Calculation

Find$\frac{db}{da}$ .

$$\begin{gathered} \frac{db}{da}=a{e}^{2a}\left(2\right)+e^{2a} \\ =2a{e}^{2a}+e^{2a} \end{gathered}$$

Find$\frac{dc}{da}$ .

$$\begin{gathered} \frac{dc}{da}=\cos \left(a-b\right)\left(1-\frac{db}{da}\right) \\ =\cos \left(a-b\right)\left(1-2a{e}^{2a}-e^{2a}\right) \\ \frac{dc}{da}=\cos \left(a-b\right)\left(1-2b-e^{2a}\right) \end{gathered}$$

Hence the$\frac{dc}{da}$ is$\cos \left(a-b\right)\left(1-2b-e^{2a}\right)$ .