What is the tangent of y=2^{x^2-3x+5},\at x=4 ?

The tangent of y=2^{x^2-3x+5},\at x=4 is y=2560ln(2)x+512-10240ln(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=2^{x^2-3x+5},\at x=4

tangent of

y = 2^{x^{2} - 3 x + 5}, at x = 4

y = 2560 ln (2) x + 512 - 10240 ln (2)

Solution steps

Find the tangent point:

(4, 512)

Find the slope of

y = 2^{x^{2} - 3 x + 5} : \frac{d y}{d x} = ln (2) \cdot 2^{x^{2} - 3 x + 5} (2 x - 3)

Compute the slope of

y = 2^{x^{2} - 3 x + 5}

x = 4 : m = 2560 ln (2)

Find the line with slope m=

2560 ln (2)

and passing through

(4, 512) : y = 2560 ln (2) x + 512 - 10240 ln (2)

y = 2560 ln (2) x + 512 - 10240 ln (2)

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

37 arctan (x)

\frac{d y}{d x}, e^{4 x} = sin (x + 5 y)

\frac{e ^{x}}{1 + x}

What is the tangent of y=2^{x^2-3x+5},\at x=4 ?
The tangent of y=2^{x^2-3x+5},\at x=4 is y=2560ln(2)x+512-10240ln(2)