What is the tangent of f(x)=sec(x),\at x=-(7pi)/4 ?

The tangent of f(x)=sec(x),\at x=-(7pi)/4 is y=sqrt(2)x+sqrt(2)+(7pi)/(2sqrt(2))

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tangent of f(x)=sec(x),\at x=-(7pi)/4

tangent of

f (x) = sec (x), at x = - \frac{7 π}{4}

y = 2 x + 2 + \frac{7 π}{2 2}

Solution steps

Find the tangent point:

(- \frac{7 π}{4}, 2)

Find the slope of

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

Compute the slope of

f (x) = sec (x)

x = - \frac{7 π}{4} : m = 2

Find the line with slope m=

2

and passing through

(- \frac{7 π}{4}, 2) : y = 2 x + 2 + \frac{7 π}{2 2}

y = 2 x + 2 + \frac{7 π}{2 2}

\int (x^{2} + 1)^{3} x d x

\frac{d}{d x} (\frac{4 x ^{2} + 6 x + 8}{x})

\int \frac{y}{1 + x ^{2}} d y

f (x) = 4 + 4 x^{2} - 2 x^{3}, at x = a

t \to 0 lim (sin (6 t))

What is the tangent of f(x)=sec(x),\at x=-(7pi)/4 ?
The tangent of f(x)=sec(x),\at x=-(7pi)/4 is y=sqrt(2)x+sqrt(2)+(7pi)/(2sqrt(2))