# How do you differentiate #f(x)=4x*(x-4)*tanx# using the product rule?

##### 2 Answers

#### Answer:

#### Explanation:

The product rule states that

I'm going to distribute the polynomial terms to take a shortcut:

Simplifying by factoring:

(Note: don't try substituting the trig identity

#### Answer:

# f'(x) = 8(x-2)tanx + 4x(x-4)sec^2x #

#### Explanation:

We have:

# f(x) = 4x(x-4)tanx #

We will apply the Product Rule for Differentiation:

# d/dx(uv)=u(dv)/dx+(du)/dxv # , or,# (uv)' = (du)v + u(dv) #

I was taught to remember the rule in words; "*The first times the derivative of the second plus the derivative of the first times the second* ".

This can be extended to three products:

# d/dx(uvw)=uv(dw)/dx+u(dv)/dxw + (du)/dxvw#

# { ("Let", u = 4x, => , (du)/dx = 4), ("And" ,v =x-4, =>, (dv)/dx = 1 ), ("And", w = tanx, =>, (dw)/dx = sec^2x ) :}#

Then:

# d/dx(uvw) = (du)/dxvw + u(dv)/dxw+ + uv(dw)/dx#

Gives us:

# f'(x) = (4)(x-4)tanx + (4x)(1)tanx+ + 4x(x-4)sec^2x#

# " " = 4{(x-4)tanx + xtanx + x(x-4)sec^2x} #

# " " = 4{2xtanx-4tanx + x(x-4)sec^2x} #

# " " = 4{(2x-4)tanx + x(x-4)sec^2x} #

# " " = 8(x-2)tanx + 4x(x-4)sec^2x #