What is (\partial)/(\partial a)((h)(b+cos(ab))) ?

The answer to (\partial)/(\partial a)((h)(b+cos(ab))) is -hbsin(ba)

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(\partial)/(\partial a)((h)(b+cos(ab)))

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

- hb sin (ba)

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial a} (b) = 0

\frac{\partial}{\partial a} (cos (ab)) = - sin (ab) b

= h (0 - sin (ab) b)

Simplify

= - hb sin (ba)

\frac{3 x ^{2}}{( x ^{3} + 1 ) ^{2}}

(1 - i 3)^{4}

(1 + x^{2}) arctan (x) - x

(1 - 2 i)^{10}

a (- 1 - 3) b (5, 5) c (6, - 2)

What is (\partial)/(\partial a)((h)(b+cos(ab))) ?
The answer to (\partial)/(\partial a)((h)(b+cos(ab))) is -hbsin(ba)