What is the tangent of f(x)=4sin(x)+3cos(x),\at x=0 ?

The tangent of f(x)=4sin(x)+3cos(x),\at x=0 is y=4x+3

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tangent of f(x)=4sin(x)+3cos(x),\at x=0

tangent of

f (x) = 4 sin (x) + 3 cos (x), at x = 0

y = 4 x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

f (x) = 4 sin (x) + 3 cos (x) : \frac{df ( x )}{d x} = 4 cos (x) - 3 sin (x)

Compute the slope of

f (x) = 4 sin (x) + 3 cos (x)

x = 0 : m = 4

Find the line with slope m=

4

and passing through

(0, 3) : y = 4 x + 3

y = 4 x + 3

(x + 1)^{^{'}}

\int \frac{2 x - 4 + 8 x}{x} d x

x \to 1 lim (\frac{x ^{2} - 3 x}{x})

(2 x y + y^{4}) d x + (3 x^{2} + 6 x y^{3}) d y = 0

y^{^{'}} + \frac{4}{5} y = 1 - \frac{1}{4} t

What is the tangent of f(x)=4sin(x)+3cos(x),\at x=0 ?
The tangent of f(x)=4sin(x)+3cos(x),\at x=0 is y=4x+3