How do you take the derivative of #tan(4x)^tan(5x)#?

Question

Evelyn Agard · Accepted Answer

#(dy)/(dx)=y{(4tan(5x))/(sin(4x)cos(4x))+5sec^2(5x)lntan(4x)}#

Let,

#y=tan(4x)^tan(5x)#

Taking natural log. ,both sides

#lny=lntan(4x)^tan(5x)#

#lny=tan(5x)*lntan(4x)#

Diff.w.r.t #x# ,Using Product Rule:

#1/y(dy)/(dx)=tan(5x)d/(dx)(lntan(4x))+lntan(4x)d/(dx)(tan(5x))#

#1/y(dy)/(dx)=tan(5x)*4/tan(4x)sec^2(4x)+lntan(4x)5sec^2(5x)#

#1/y(dy)/(dx)=4tan(5x)cos(4x)/sin(4x)xx1/cos^2(4x)+5sec^2(5x)lnt an(4x)#

#1/y(dy)/(dx)=(4tan(5x))/(sin(4x)cos(4x))+5sec^2(5x)lntan(4x)#

#(dy)/(dx)=y{(4tan(5x))/(sin(4x)cos(4x))+5sec^2(5x)lntan(4x)}#

Charles Anctil · Answer

undefined To find the derivative of $ \tan(4x)^{\tan(5x)} $, you would use the chain rule and the properties of logarithmic differentiation. First, rewrite the function using logarithms:

$$ y = \tan(4x)^{\tan(5x)} $$
$$ \ln(y) = \ln(\tan(4x)^{\tan(5x)}) $$
$$ \ln(y) = \tan(5x) \cdot \ln(\tan(4x)) $$

Now, take the derivative of both sides with respect to $ x $:

$$ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}[\tan(5x) \cdot \ln(\tan(4x))] $$

Apply the product rule on the right side:

$$ \frac{1}{y} \cdot \frac{dy}{dx} = \sec^2(5x) \cdot \ln(\tan(4x)) + \tan(5x) \cdot \frac{d}{dx}[\ln(\tan(4x))] $$

Now, find the derivative of $ \ln(\tan(4x)) $ using the chain rule:

$$ \frac{1}{y} \cdot \frac{dy}{dx} = \sec^2(5x) \cdot \ln(\tan(4x)) + \tan(5x) \cdot \frac{1}{\tan(4x)} \cdot \sec^2(4x) \cdot 4 $$

Finally, solve for $ \frac{dy}{dx} $ by multiplying both sides by $ y $:

$$ \frac{dy}{dx} = y \left( \sec^2(5x) \cdot \ln(\tan(4x)) + \tan(5x) \cdot \frac{1}{\tan(4x)} \cdot \sec^2(4x) \cdot 4 \right) $$

$$ \frac{dy}{dx} = \tan(4x)^{\tan(5x)} \left( \sec^2(5x) \cdot \ln(\tan(4x)) + \tan(5x) \cdot \frac{1}{\tan(4x)} \cdot \sec^2(4x) \cdot 4 \right) $$

That's the derivative of $ \tan(4x)^{\tan(5x)} $.