How do I find the derivative of #y = (sin(e^x))^ln(x^2)#?

Question

Isabella Ackerman · Accepted Answer

#color(blue)(y'=(sin e^x)^(ln x^2)[e^x*ln x^2*cot e^x+2/x*ln sin e^x])#

The given equation is

#y=(sin e^x)^(ln x^2)#

Take the natural logarithm of both sides then differentiate with respect to x

#y=(sin e^x)^(ln x^2)#

#ln y=ln (sin e^x)^(ln x^2)#

#ln y=(ln x^2)*ln (sin e^x)#

We now have a derivative of a product at the right sides of the equation.

Obtain the derivative of both sides

#ln y=(ln x^2)*ln (sin e^x)#

#d/dx(ln y)=d/dx[(ln x^2)*ln (sin e^x)]#

#1/y*y'=(ln x^2)*d/dx(ln sin e^x)+(ln sin e^x)*d/dx(ln x^2)#

#1/y*y'=(ln x^2)*1/( sin e^x)*d/dx(sin e^x)+(ln sin e^x)*1/x^2*d/dx(x^2)#

#1/y*y'=(ln x^2)*1/( sin e^x)(cos e^x)d/dx(e^x)+(ln sin e^x)*1/x^2*2x#

#1/y*y'=(ln x^2)*1/( sin e^x)(cos e^x)(e^x)+2/x*ln sin e^x#

#1/y*y'=(ln x^2)*(cot e^x)(e^x)+2/x*ln sin e^x#

Multiply both sides by #y# then simplify

#y*1/y*y'=y*(ln x^2)*(cot e^x)(e^x)+2/x*ln sin e^x#

#y'=y*(ln x^2)*(cot e^x)(e^x)+2/x*ln sin e^x#

#y'=(sin e^x)^(ln x^2)[e^x*ln x^2*cot e^x+2/x*ln sin e^x]#

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

Isabella Ackerman · Answer

undefined To find the derivative of $y = (\sin(e^x))^{\ln(x^2)}$, you can use the chain rule and the properties of logarithmic and exponential functions. Here's the step-by-step process:

1. Take the natural logarithm of both sides of the equation: 
$$ \ln(y) = \ln\left((\sin(e^x))^{\ln(x^2)}\right) $$

2. Use the property $\ln(a^b) = b\ln(a)$ to simplify:
$$ \ln(y) = \ln(x^2)\ln(\sin(e^x)) $$

3. Differentiate both sides with respect to $x$:
$$ \frac{1}{y} \frac{dy}{dx} = \frac{2}{x} \ln(\sin(e^x)) + \frac{x^2}{\sin(e^x)} \cdot \frac{\cos(e^x)e^x}{\sin(e^x)} $$

4. Solve for $\frac{dy}{dx}$:
$$ \frac{dy}{dx} = y \left(\frac{2}{x} \ln(\sin(e^x)) + \frac{x^2}{\sin(e^x)} \cdot \frac{\cos(e^x)e^x}{\sin(e^x)}\right) $$

5. Substitute back $y = (\sin(e^x))^{\ln(x^2)}$:
$$ \frac{dy}{dx} = (\sin(e^x))^{\ln(x^2)} \left(\frac{2}{x} \ln(\sin(e^x)) + \frac{x^2}{\sin(e^x)} \cdot \frac{\cos(e^x)e^x}{\sin(e^x)}\right) $$

That's the derivative of $y$ with respect to $x$.