What is (\partial)/(\partial {b)}(e^asin(a+{b})) ?

The answer to (\partial)/(\partial {b)}(e^asin(a+{b})) is e^acos(a+{b})

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(\partial)/(\partial {b)}(e^asin(a+{b}))

\frac{\partial}{\partial b} (e^{a} sin (a + b))

e^{a} cos (a + b)

Solution steps

\frac{\partial}{\partial b} (e^{a} sin (a + b))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= e^{a} \frac{\partial}{\partial b} (sin (a + b))

Apply the chain rule:

cos (a + b) \frac{\partial}{\partial b} (a + b)

= cos (a + b) \frac{\partial}{\partial b} (a + b)

\frac{\partial}{\partial b} (a + b) = 1

= e^{a} cos (a + b) \cdot 1

Simplify

= e^{a} cos (a + b)

\int x^{2} - 4 x + 40 d x

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = 3

f (x) = (12 x^{3} - 9 x)^{\frac{4}{3}}

\frac{d}{d x} (x^{3} - 9 x)

\int - \frac{2 x}{1 + x ^{2}} d x

What is (\partial)/(\partial {b)}(e^asin(a+{b})) ?
The answer to (\partial)/(\partial {b)}(e^asin(a+{b})) is e^acos(a+{b})