en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

derivative of

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

area e^x,xe^x,0 area

e^{x}, x e^{x}, 0

(\partial)/(\partial x)(x^2+y^2+z^2-25)

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2} - 25)

y^'+2y=2x+e^{4x}

y^{^{'}} + 2 y = 2 x + e^{4 x}

integral of 2xcos(9x)

\int 2 x cos (9 x) d x

tangent of f(x)= 3/(x-2),\at x=3 tangent of

f (x) = \frac{3}{x - 2}, at x = 3

Frequently Asked Questions (FAQ)

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