Question

find the slope of the tangent to the curve y=5x^2 at (-1,5)​

Answer 1

$\textbf{Given:}$

$\textsf{Curve is}\;\mathsf{y=5\,x^2}$

$\textbf{To find:}$

$\textsf{Slope of tangent to the given curve at (-1,5)}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{y=5\,x^2}$

$\textsf{Differentiate with respect to 'x'}$

$\mathsf{\dfrac{dy}{dx}=5(2x)}$

$\mathsf{\dfrac{dy}{dx}=10x}$

$\mathsf{Slope\;of\;tangent\;to\;the\;curve}$

$\mathsf{=\left(\dfrac{dy}{dx}\right)_{(-1,5)}}$

$\mathsf{=10(-1)}$

$\mathsf{=-10}$

$\implies\boxed{\mathsf{Slope\;of\;tangent\;to\;the\;curve=-10}}$

$\textbf{Find more:}$

Equation of the tangent to the curve y = 1-e^-x/2 at the point where the curve cuts y axis is

https://brainly.in/question/6188805

Answer 2

Answer:

-10

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