What is the tan(f(x))(x)=-4x+1/2 x-5,\at x=1 ?

The tan(f(x))(x)=-4x+1/2 x-5,\at x=1 is tan(f(x))x\circ \at x

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tan(f(x))(x)=-4x+1/2 x-5,\at x=1

tan (f (x)) (x) = - 4 x + \frac{1}{2} x - 5, at x = 1

tan (f (x)) x \circ at x

Solution steps

Find

tan (f (x)) (x) \circ at x

given

tan (f (x)) (x) (x) = - 4 x + \frac{1}{2} x - 5, at x = 1

= tan (f (x)) x \circ at x

\frac{\partial}{\partial x} (3 x^{2 y})

s (t) = - 9.8 t^{2} + 2 t + 1

y = (2 x + 3)^{5} + 10 x (2 x + 3)^{4}

\int e x^{2} + π d x

f (x) = arcsin^{2} (x)

What is the tan(f(x))(x)=-4x+1/2 x-5,\at x=1 ?
The tan(f(x))(x)=-4x+1/2 x-5,\at x=1 is tan(f(x))x\circ \at x