Derivatives Evaluate the following derivatives. ddy(ysin⁡y)

Question

Derivatives Evaluate the following derivatives.  ddy(ysin⁡y)

Thomas White · Accepted Answer

Step 1  Given- ddy(ysin⁡y)  To evaluate- The above derivative.  Step 2  Explanation- Rewrite the given expression as,  z=(ysin⁡y)  Now,taking log both sides, we get,  ln⁡z=s∈yln⁡y  Differetiating both sides, we get,  1z⋅dzdy=cos⁡y⋅ln⁡y+sin⁡y⋅1y  dzdy=z(ln⁡ycos⁡y+sin⁡yy)  dzdy=ysin⁡y(ln⁡ycos⁡y+sin⁡yy)  Answer-Hence, the derivative of the expression ddy(ysin⁡y) is ysin⁡y(ln⁡ycos⁡y+sin⁡yy)

servidopolisxv · Accepted Answer

To evaluate  ddy(ysin⁡y)=ddy(eln⁡ysin⁡y)  The function eln⁡ysin⁡y is the compostion f(g(y)) of two function f(u)=eu and g(y)=ln⁡ysin⁡y. According to the chain rule  ddy(f(g(y)))=ddu(f(u))ddy(g(y))  we get  ddy(eln⁡ysin⁡y)=ddu(eu)ddy(ln⁡ysin⁡y)  =eu(ddy(ln⁡ysin⁡y)+ddy(sin⁡y)ln⁡y)  =ysin⁡y(1ysin⁡y+cos⁡yln⁡y)  =ysin⁡y−1(ycos⁡yln⁡y+sin⁡y)  Hence,  ddy(ysin⁡y)=ysin⁡y−1(ycos⁡yln⁡y+sin⁡y)  Result:  ddy(ysin⁡y)=ysin⁡y−1(ycos⁡yln⁡y+sin⁡y)

Derivatives Evaluate the following derivatives. \frac{d}{dy}(y^{\sin y})

Answered question

Answer & Explanation

New Questions in Calculus 1