Equation of Tangent

[Jason76]

Product Rule:

Given: \(\displaystyle f(x) g(x)\)

\(\displaystyle g(x)[f'(x)] + f(x)[g'(x)]\)

Find Tangent line with point \(\displaystyle (0,5)\)

\(\displaystyle f(x) = 5e^{x} \cos(x)\)

\(\displaystyle f'(x) = \cos(x) [\dfrac{d}{dx} 5e^{x}] + 5e^{x} [\dfrac{d}{dx} \cos(x)] \)

\(\displaystyle f'(x) = \cos(x) [5e^{x}] + 5e^{x} [-\sin(x))] \)

\(\displaystyle f'(x) = \cos(x) 5e^{x} + [-5e^{x} \sin(x)]\)

\(\displaystyle f'(x) = \cos(x) 5e^{x} - 5e^{x} \sin(x)]\)

\(\displaystyle f'(x) = 5e^{x}[\cos(x) - \sin(x)]\)

\(\displaystyle f'(0) = 5e^{0}[\cos(0) - \sin(0)]\)

\(\displaystyle f'(0) = 5(1 - 0) = 5\)

\(\displaystyle y - 5 = 5(x - 0)\)

\(\displaystyle y - 5 = 5x - 0\)

\(\displaystyle y = 5x - 5\) Answer

 

[lookagain]

Product Rule:

Given: \(\displaystyle f(x) g(x)\)

\(\displaystyle g(x)[f'(x)] + f(x)[g'(x)]\)

Find Tangent line with point \(\displaystyle (0,5)\)

\(\displaystyle f(x) = 5e^{x} \cos(x)\)

\(\displaystyle f'(x) = \cos(x) [\dfrac{d}{dx} 5e^{x}] + 5e^{x} [\dfrac{d}{dx} \cos(x)] \)

\(\displaystyle f'(x) = \cos(x) [5e^{x}] + 5e^{x} [-\sin(x))] \)

\(\displaystyle f'(x) = \cos(x) 5e^{x} + [-5e^{x} \sin(x)]\)

\(\displaystyle f'(x) = \cos(x) 5e^{x} - 5e^{x} \sin(x)] \ \ \ \ \)(Get rid of that close parenthesis that isn't paired with anything.)

\(\displaystyle f'(x) = 5e^{x}[\cos(x) - \sin(x)]\)

\(\displaystyle f'(0) = 5e^{0}[\cos(0) - \sin(0)]\)

\(\displaystyle f'(0) = 5(1 - 0) = 5\)

\(\displaystyle y - 5 = 5(x - 0)\)

\(\displaystyle y - 5 = 5x - 0\)

\(\displaystyle y = 5x - 5\) Answer\(\displaystyle \ \ \ \) <------ After all that, you messed up! I bet you made that error from doing it in your head, right?

So, what should be the answer?



Edit: You can't call the original function "f(x)" and part of (a factor of) the original function also "f(x)."