How do you differentiate y= x^(11cosx)y=x11cosx?
1 Answer
Explanation:
y=x^(11cos(x))y=x11cos(x)
To deal with tricky exponents like this, let's take the natural logarithm of both sides and remember the rule
ln(y)=ln(x^(11cos(x)))ln(y)=ln(x11cos(x))
ln(y)=11cos(x)ln(x)ln(y)=11cos(x)ln(x)
Now take the derivative on both sides. On the left, we'll need the chain rule. On the right, we'll use the product rule.
1/y(dy/dx)=11(d/dxcos(x))ln(x)+11cos(x)(d/dxln(x))1y(dydx)=11(ddxcos(x))ln(x)+11cos(x)(ddxln(x))
1/y(dy/dx)=11(-sin(x))ln(x)+11cos(x)(1/x)1y(dydx)=11(−sin(x))ln(x)+11cos(x)(1x)
Solving for the derivative:
dy/dx=y((11cos(x))/x-11sin(x)ln(x))dydx=y(11cos(x)x−11sin(x)ln(x))
dy/dx=11x^(11cos(x))(cos(x)/x-sin(x)ln(x))dydx=11x11cos(x)(cos(x)x−sin(x)ln(x))
