Algebra
Calculus
Trigonometry
Matrix
Question
Find the partial derivative
y2+2ycos(x)+cos2(x)2ysin(x)
Evaluate
∂x∂(y+cos(x)2y)
Use differentiation rule ∂x∂(g(x)f(x))=(g(x))2∂x∂(f(x))×g(x)−f(x)×∂x∂(g(x))
(y+cos(x))2∂x∂(2y)(y+cos(x))−2y×∂x∂(y+cos(x))
Use ∂x∂(c)=0 to find derivative
(y+cos(x))20×(y+cos(x))−2y×∂x∂(y+cos(x))
Evaluate
More Steps
Evaluate
∂x∂(y+cos(x))
Use differentiation rule ∂x∂(f(x)±g(x))=∂x∂(f(x))±∂x∂(g(x))
∂x∂(y)+∂x∂(cos(x))
Use ∂x∂(c)=0 to find derivative
0+∂x∂(cos(x))
Use ∂x∂(cosx)=−sinx to find derivative
0−sin(x)
Removing 0 doesn't change the value,so remove it from the expression
−sin(x)
(y+cos(x))20×(y+cos(x))−2y(−sin(x))
Any expression multiplied by 0 equals 0
(y+cos(x))20−2y(−sin(x))
Evaluate
(y+cos(x))20−(−2ysin(x))
Evaluate
More Steps
Evaluate
0−(−2ysin(x))
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
0+2ysin(x)
Removing 0 doesn't change the value,so remove it from the expression
2ysin(x)
(y+cos(x))22ysin(x)
Rewrite the expression
2y(y+cos(x))−2sin(x)
Rewrite the expression
2y(y2+2ycos(x)+cos2(x))−1sin(x)
Solution
y2+2ycos(x)+cos2(x)2ysin(x)
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