What is the tangent of 8sin(x)+6cos(x),\at x=0 ?

The tangent of 8sin(x)+6cos(x),\at x=0 is y=8x+6

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tangent of 8sin(x)+6cos(x),\at x=0

tangent of

8 sin (x) + 6 cos (x), at x = 0

y = 8 x + 6

Solution steps

Find the tangent point:

(0, 6)

Find the slope of

y = 8 sin (x) + 6 cos (x) : \frac{d y}{d x} = 8 cos (x) - 6 sin (x)

Compute the slope of

y = 8 sin (x) + 6 cos (x)

x = 0 : m = 8

Find the line with slope m=

8

and passing through

(0, 6) : y = 8 x + 6

y = 8 x + 6

\frac{d}{d x} (x sec^{2} (x) + tan (x))

y = x e^{- 0.1 x}, y = 0, x = 3

\int cos^{3} (7 - x) sin (7 - x) d x

y^{^{'}} - x^{3} = x^{3} y

f (x) = \frac{x}{x + 9}, at x = 1

What is the tangent of 8sin(x)+6cos(x),\at x=0 ?
The tangent of 8sin(x)+6cos(x),\at x=0 is y=8x+6