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

Question

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

Isaiah Allen · Accepted Answer

#y'=(3e^(3x))/(2*cos(x+2y))-1/2#

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

#e^(3x)=sin(x+2y)#

#d/dx(e^(3x))=d/dx(sin(x+2y))#

#e^(3x)d/dx(3x)=cos(x+2y)*d/dx(x+2y)#

#(e^(3x))(3)=cos(x+2y)(1+2y')#

#(3e^(3x))/(cos(x+2y))=(cancelcos(x+2y)(1+2y'))/(cancelcos(x+2y))#

#(3e^(3x))/(cos(x+2y))=1+2y'#

#(3e^(3x))/(cos(x+2y))-1=2y'#

#((3e^(3x))/(cos(x+2y))-1)/2=(2y')/2#

#((3e^(3x))/(cos(x+2y))-1)/2=(cancel2y')/cancel2#

#y'=(3e^(3x))/(2*cos(x+2y))-1/2#

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

Samantha Adkison · Answer

undefined To implicitly differentiate $ e^{3x} = \sin(x+2y) $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Use the chain rule and product rule where necessary.
3. Solve for $ \frac{dy}{dx} $ in terms of $ x $ and $ y $.

The implicit differentiation of the given equation $ e^{3x} = \sin(x+2y) $ yields:

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

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

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

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