Find an equation of the tangent line to the graph of $y = e ^ { 5 x }$ at the point (0,1) .
A) $y = x + 1$
B) $y = \ln ( 5 ) x + 1$
C) $y = 6 x + 1$
D) $y = 5 x + 1$
E) $y = 5 x - 1$

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to the graph of $y = e^{5x}$ at the point (0,1), we first need to find the derivative of $y$ with respect to $x$, which gives us the slope of the tangent line at any point $x$. The derivative of $y = e^{5x}$ with respect to $x$ is $y' = 5e^{5x}$. At $x = 0$, $y' = 5e^{5*0} = 5$. Therefore, the slope of the tangent line at the point (0,1) is 5. The equation of a line with slope $m$ passing through a point $(x_1, y_1)$ is given by $y - y_1 = m(x - x_1)$. Substituting $m = 5$, $x_1 = 0$, and $y_1 = 1$, we get $y - 1 = 5(x - 0)$, which simplifies to $y = 5x + 1$.

Find an Equation of the Tangent Line to the Graph y=e5xy = e ^ { 5 x }y=e5x

Find an Equation of the Tangent Line to the Graph $y = e ^ { 5 x }$