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Popular >

(f(t)\circ g(t))'

Solution

(f (t) \circ g (t))^{'}

Solution

\circ (f' (t) g (t) + g' (t) f (t))

Solution steps

(f (t) \circ g (t))^{'}

Take the constant out:

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

= \circ (f (t) g (t))^{^{'}}

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = f (t), g = g (t)

= \circ (f^{^{'}} (t) g (t) + g^{^{'}} (t) f (t))

= \circ (f^{^{'}} (t) g (t) + g^{^{'}} (t) f (t))

Popular Examples

integral from 2 to 3 of (x+1)/(x^2-x)

\int_{2}^{3} \frac{x + 1}{x ^{2} - x} d x

limit as x approaches 1 of (3x+2)/(4x-5)

x \to 1 lim (\frac{3 x + 2}{4 x - 5})

(\partial)/(\partial x)(10-(3x^2+3y^2))

\frac{\partial}{\partial x} (10 - (3 x^{2} + 3 y^{2}))

derivative of 3x^2-4x+1

\frac{d}{d x} (3 x^{2} - 4 x + 1)

integral of 3/(e^{t/2)}

\int \frac{3}{e ^{\frac{t}{2}}} d t

Frequently Asked Questions (FAQ)

What is the (f(t)\circ g(t))' ?
The (f(t)\circ g(t))' is \circ (f'(t)g(t)+g'(t)f(t))
What is the first (f(t)\circ g(t))' ?
The first (f(t)\circ g(t))' is \circ (f'(t)g(t)+g'(t)f(t))