#y=sin^2(lnx^2)# What is #dy/dx# ?

#y=sin^2(lnx^2)# . Find #dy/dx#

2 Answers
Apr 20, 2018

Answer:

#dy/dx = (4sin(lnx^2)cos(lnx^2))/x#

Explanation:

If #y=sin^2(lnx^2)# then

#dy/dx=d/(dlnx^2)sin^2(lnx^2) d/dxlnx^2=2sin(lnx^2)cos(lnx^2)(2x)/x^2=(4sin(lnx^2)cos(lnx^2))/x#

Apr 21, 2018

Answer:

#dy/dx=(4sin(ln(x^2))cos(ln(x^2)))/x#

Explanation:

We use the chain rule twice here:

#d/dx[f(g(x))]=f'(g(x))*g'(x)#

#=>dy/dx=d/dx[sin^2(ln(x^2))]#

Some rules to recall here:

#d/dx[x^n]=nx^(n-1)# if #n# is a constant.

#d/dx[ln(x)]=1/x#

#d/dx[sinx]=cosx#

#=>dy/dx=2*sin(ln(x^2))*d/dx[sin(ln(x^2))]#

#=>dy/dx=2*sin(ln(x^2))*cos(ln(x^2))*d/dx[ln(x^2)]#

#=>dy/dx=2*sin(ln(x^2))*cos(ln(x^2))*1/(x^2)*d/dx[x^2]#

#=>dy/dx=2*sin(ln(x^2))*cos(ln(x^2))*1/(x^2)*2x#

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

#=>dy/dx=(4sin(ln(x^2))cos(ln(x^2)))/x#

That is our answer!