What is the tangent of-9x^{1/3}+5,\at x=4 ?

The tangent of-9x^{1/3}+5,\at x=4 is y=-3/(4^{2/3)}x-9*4^{1/3}+5+3*2^{2/3}

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tangent of-9x^{1/3}+5,\at x=4

tangent of

- 9 x^{\frac{1}{3}} + 5, at x = 4

y = - \frac{3}{4 ^{\frac{2}{3}}} x - 9 \cdot 4^{\frac{1}{3}} + 5 + 3 \cdot 2^{\frac{2}{3}}

Solution steps

Find the tangent point:

(4, - 9 \cdot 4^{\frac{1}{3}} + 5)

Find the slope of

y = - 9 x^{\frac{1}{3}} + 5 : \frac{d y}{d x} = - \frac{3}{x ^{\frac{2}{3}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{3}{4 ^{\frac{2}{3}}}

Find the line with slope m=

- \frac{3}{4 ^{\frac{2}{3}}}

and passing through

(4, - 9 \cdot 4^{\frac{1}{3}} + 5) : y = - \frac{3}{4 ^{\frac{2}{3}}} x - 9 \cdot 4^{\frac{1}{3}} + 5 + 3 \cdot 2^{\frac{2}{3}}

y = - \frac{3}{4 ^{\frac{2}{3}}} x - 9 \cdot 4^{\frac{1}{3}} + 5 + 3 \cdot 2^{\frac{2}{3}}

\frac{\partial}{\partial z} (ln (x + z) - yz)

64 y^{^{''}} + 80 y^{^{'}} + 25 y = 0

y^{^{''}} + 4 y = x

\int \frac{4 x ^{3} + 2}{x ^{2}} d x

\frac{d ^{2}}{d x ^{2}} (sin (5 x))

What is the tangent of-9x^{1/3}+5,\at x=4 ?
The tangent of-9x^{1/3}+5,\at x=4 is y=-3/(4^{2/3)}x-9*4^{1/3}+5+3*2^{2/3}