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Popular >

derivative of f(t)=e^{5tsin(2t)}

Solution

derivative of

f (t) = e^{5 t s i n (2 t)}

Solution

e^{5 t s i n (2 t)} \cdot 5 (sin (2 t) + 2 t cos (2 t))

Solution steps

\frac{d}{d t} (e^{5 t s i n (2 t)})

Apply the chain rule:

e^{5 t s i n (2 t)} \frac{d}{d t} (5 t sin (2 t))

= e^{5 t s i n (2 t)} \frac{d}{d t} (5 t sin (2 t))

\frac{d}{d t} (5 t sin (2 t)) = 5 (sin (2 t) + 2 t cos (2 t))

= e^{5 t s i n (2 t)} \cdot 5 (sin (2 t) + 2 t cos (2 t))

Popular Examples

limit as x approaches 2 of 3x^4+4x^3-x+2

x \to 2 lim (3 x^{4} + 4 x^{3} - x + 2)

4y^{''}-4y^'+13y=0

4 y^{^{''}} - 4 y^{^{'}} + 13 y = 0

implicit (dy)/(dx),y=e^{4x^2+1}implicit derivative

\frac{d y}{d x}, y = e^{4 x^{2} + 1}

(\partial)/(\partial y)(yz)

\frac{\partial}{\partial y} (yz)

derivative of sqrt(x)-2sqrt(y)

\frac{d}{d x} (x - 2 y)

Frequently Asked Questions (FAQ)

What is the derivative of f(t)=e^{5tsin(2t)} ?
The derivative of f(t)=e^{5tsin(2t)} is e^{5tsin(2t)}*5(sin(2t)+2tcos(2t))
What is the first derivative of f(t)=e^{5tsin(2t)} ?
The first derivative of f(t)=e^{5tsin(2t)} is e^{5tsin(2t)}*5(sin(2t)+2tcos(2t))