What will be the differentiation of this?
d/dx (6tan(3x)sec ^2 (3x))

Question

What will be the differentiation of this? 
d/dx  (6tan(3x)sec ^2 (3x))

Christopher Agresta · Accepted Answer

#(dy)/(dx)=18sec^2 3x(2tan^2 3x+sec^2 3x)#

Let ,

#y= 6tan(3x)sec ^2 (3x)#

We take,

#color(red)(y=u*v ,where, u=6tan3x and v=sec^2 3x#

Using chain rule:

#=>(du)/(dx)=6sec^2 3xd/(dx)(3x)=6sec^2 3x*3=18sec^2 3x#

#and (dv)/(dx)=2sec 3xd/(dx)(sec3x)=2sec3x*sec3xtan3x*3#

#=>(dv)/(dx)=6sec^2 3xtan3x#

Diff.w.r.t. #x# using product rule:

#color(blue)(d/(dx)(u*v)=u*(dv)/(dx)+v*(du)/(dx)#

#:.(dy)/(dx)=6tan3x(6sec^2 3xtan3x)+sec^2 3x(18sec^2 3x)#

#(dy)/(dx)=36sec^2 3xtan^2 3x+18sec^4 3x#

#(dy)/(dx)=18sec^2 3x(2tan^2 3x+sec^2 3x)#

Samantha Adkison · Answer

undefined To differentiate $ 6\tan(3x)\sec^2(3x) $ with respect to $ x $, you can apply the product rule. The product rule states that if you have two functions $ u(x) $ and $ v(x) $, then the derivative of their product is given by:

$$ \frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x) $$

Where $ u'(x) $ and $ v'(x) $ represent the derivatives of $ u(x) $ and $ v(x) $ with respect to $ x $ respectively. Applying this to the given function:

$$ u(x) = 6\tan(3x) $$
$$ v(x) = \sec^2(3x) $$

Taking derivatives:
$$ u'(x) = 6\cdot3\sec^2(3x) $$
$$ v'(x) = 2\sec(3x)\tan(3x)\cdot3 $$

Substituting into the product rule formula:
$$ \frac{d}{dx}(6\tan(3x)\sec^2(3x)) = (6\cdot3\sec^2(3x))(\sec^2(3x)) + (6\tan(3x))(2\sec(3x)\tan(3x)\cdot3) $$

Simplify this expression to get the final result.

What will be the differentiation of this? d/dx (6tan(3x)sec ^2 (3x))