What is the tangent of f(x)=(e^{5x}-4)^2,(0,9) ?

The tangent of f(x)=(e^{5x}-4)^2,(0,9) is y=-30x+9

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tangent of f(x)=(e^{5x}-4)^2,(0,9)

tangent of

f (x) = (e^{5 x} - 4)^{2}, (0, 9)

y = - 30 x + 9

Solution steps

Find the slope of

f (x) = (e^{5 x} - 4)^{2} : \frac{df ( x )}{d x} = 10 e^{5 x} (e^{5 x} - 4)

Compute the slope of

f (x) = (e^{5 x} - 4)^{2}

(0, 9) : m = - 30

Find the line with slope m=

- 30

and passing through

(0, 9) : y = - 30 x + 9

y = - 30 x + 9

\frac{d}{d x} (sin (x) - x \cdot cos (x))

\frac{d y}{d x} + y cos (x) = sin (x) cos (x)

- csc (x) - sin (x)

- 6 t^{3}

f (x) = x^{3} + 6 x^{2} + x

What is the tangent of f(x)=(e^{5x}-4)^2,(0,9) ?
The tangent of f(x)=(e^{5x}-4)^2,(0,9) is y=-30x+9