How do you differentiate #y=ln ln(2x^4)#?

Question

6038 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-15T16:49:03

#dy/dx=4/(xln(2x^4))#

#y=ln(ln(2x^4))#

Through the chain rule, we see that:

#d/dxln(color(red)f(x))=1/color(red)f(x)*d/dxcolor(red)f(x)#

So, here, we see that:

#dy/dx=d/dxln(color(red)ln(2x^4))=1/color(red)ln(2x^4)*d/dxcolor(red)ln(2x^4)#

Note that we will use the same rule again to find #d/dxln(2x^4)#:

#dy/dx=1/ln(2x^4) * [d/dxln(color(red)(2x^4))]=1/ln(2x^4) * [1/color(red)(2x^4)*d/dxcolor(red)(2x^4)]#

Differentiating #2x^4# through the power rule gives #8x^3#:

#dy/dx=1/ln(2x^4) * 1/(2x^4) * 8x^3#

#dy/dx=1/ln(2x^4) * 1/x * 4#

#dy/dx=4/(xln(2x^4))#