# How do you implicitly differentiate  e^(3x)=sin(x+2y) ?

$y ' = \frac{3 {e}^{3 x}}{2 \cdot \cos \left(x + 2 y\right)} - \frac{1}{2}$

#### Explanation:

From the given equation, just differentiate both sides of the equation with respect to $x$

${e}^{3 x} = \sin \left(x + 2 y\right)$

$\frac{d}{\mathrm{dx}} \left({e}^{3 x}\right) = \frac{d}{\mathrm{dx}} \left(\sin \left(x + 2 y\right)\right)$

${e}^{3 x} \frac{d}{\mathrm{dx}} \left(3 x\right) = \cos \left(x + 2 y\right) \cdot \frac{d}{\mathrm{dx}} \left(x + 2 y\right)$

$\left({e}^{3 x}\right) \left(3\right) = \cos \left(x + 2 y\right) \left(1 + 2 y '\right)$

$\frac{3 {e}^{3 x}}{\cos \left(x + 2 y\right)} = \frac{\cancel{\cos} \left(x + 2 y\right) \left(1 + 2 y '\right)}{\cancel{\cos} \left(x + 2 y\right)}$

$\frac{3 {e}^{3 x}}{\cos \left(x + 2 y\right)} = 1 + 2 y '$

$\frac{3 {e}^{3 x}}{\cos \left(x + 2 y\right)} - 1 = 2 y '$

$\frac{\frac{3 {e}^{3 x}}{\cos \left(x + 2 y\right)} - 1}{2} = \frac{2 y '}{2}$

$\frac{\frac{3 {e}^{3 x}}{\cos \left(x + 2 y\right)} - 1}{2} = \frac{\cancel{2} y '}{\cancel{2}}$

$y ' = \frac{3 {e}^{3 x}}{2 \cdot \cos \left(x + 2 y\right)} - \frac{1}{2}$

God bless....I hope the explanation is useful.