What is the tangent of y=f(x),\at x=1 ?

The tangent of y=f(x),\at x=1 is y= d/(d1)(f(1))x+f(1)-d/(d1)(f(1))

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tangent of y=f(x),\at x=1

tangent of

y = f (x), at x = 1

y = \frac{d}{d 1} (f (1)) x + f (1) - \frac{d}{d 1} (f (1))

Solution steps

Find the tangent point:

(1, f (1))

Find the slope of

y = f (x) : \frac{d y}{d x} = \frac{df}{d x}

Compute the slope of

y = f (x)

x = 1 : m = \frac{d}{d 1} (f (1))

Find the line with slope m=

\frac{d}{d 1} (f (1))

and passing through

(1, f (1)) : y = \frac{d}{d 1} (f (1)) x + f (1) - \frac{d}{d 1} (f (1))

y = \frac{d}{d 1} (f (1)) x + f (1) - \frac{d}{d 1} (f (1))

\int_{0}^{1} \frac{37}{x ^{5}} d x

2 y^{^{''}} + 8 y^{^{'}} + 80 y - 60 cos (8 t) = 0

\int \frac{5 ^{x}}{x} d x

\frac{d y}{d x}, x y = 20

\frac{d}{d x} (\frac{4 x}{3 ( x ^{2} - 4 ) ^{\frac{1}{3}}})

What is the tangent of y=f(x),\at x=1 ?
The tangent of y=f(x),\at x=1 is y= d/(d1)(f(1))x+f(1)-d/(d1)(f(1))