What is taylor e^{5x+5},0 ?

The answer to taylor e^{5x+5},0 is e^5+5e^5x+(25e^5)/2 x^2+(125e^5)/6 x^3+(625e^5)/(24)x^4+\ldots

Popular >

taylor e^{5x+5},0

taylor

e^{5 x + 5}, 0

e^{5} + 5 e^{5} x + \frac{25 e ^{5}}{2} x^{2} + \frac{125 e ^{5}}{6} x^{3} + \frac{625 e ^{5}}{24} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= e^{5} + \frac{\frac{d}{d x} ( e ^{5 x + 5} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{5 x + 5} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{5 x + 5} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= e^{5} + \frac{5 e ^{5}}{1 !} x + \frac{25 e ^{5}}{2 !} x^{2} + \frac{125 e ^{5}}{3 !} x^{3} + \frac{625 e ^{5}}{4 !} x^{4} + \dots

Refine

= e^{5} + 5 e^{5} x + \frac{25 e ^{5}}{2} x^{2} + \frac{125 e ^{5}}{6} x^{3} + \frac{625 e ^{5}}{24} x^{4} + \dots

\frac{d y}{d x} = (1 - 2 x) y^{2}

\int_{0}^{2} 2 π x (3 + \frac{x ^{2}}{3}) d x

f (x) = \frac{sin ( x ) + cos ( x )}{sin ( x ) - cos ( x )}

(1 + x^{3}) d y - x^{2} y d x = 0

x d y + y ln (y) d x = 0

What is taylor e^{5x+5},0 ?
The answer to taylor e^{5x+5},0 is e^5+5e^5x+(25e^5)/2 x^2+(125e^5)/6 x^3+(625e^5)/(24)x^4+\ldots