What is the tangent of y=(1+4x)^{3/2} ?

The tangent of y=(1+4x)^{3/2} is y=6(4a_{0}+1)^{1/2}x+(1+4a_{0})^{3/2}-6(4a_{0}+1)^{1/2}a_{0}

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tangent of y=(1+4x)^{3/2}

tangent of

y = (1 + 4 x)^{\frac{3}{2}}

y = 6 (4 a_{0} + 1)^{\frac{1}{2}} x + (1 + 4 a_{0})^{\frac{3}{2}} - 6 (4 a_{0} + 1)^{\frac{1}{2}} a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (1 + 4 a_{0})^{\frac{3}{2}})

Find the slope of

y = (1 + 4 x)^{\frac{3}{2}} : \frac{d y}{d x} = 6 (4 x + 1)^{\frac{1}{2}}

Compute the slope of

y = (1 + 4 x)^{\frac{3}{2}}

x = a_{0} : m = 6 (4 a_{0} + 1)^{\frac{1}{2}}

Find the line with slope m=

6 (4 a_{0} + 1)^{\frac{1}{2}}

and passing through

(a_{0}, (1 + 4 a_{0})^{\frac{3}{2}}) : y = 6 (4 a_{0} + 1)^{\frac{1}{2}} x + (1 + 4 a_{0})^{\frac{3}{2}} - 6 (4 a_{0} + 1)^{\frac{1}{2}} a_{0}

y = 6 (4 a_{0} + 1)^{\frac{1}{2}} x + (1 + 4 a_{0})^{\frac{3}{2}} - 6 (4 a_{0} + 1)^{\frac{1}{2}} a_{0}

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What is the tangent of y=(1+4x)^{3/2} ?
The tangent of y=(1+4x)^{3/2} is y=6(4a_{0}+1)^{1/2}x+(1+4a_{0})^{3/2}-6(4a_{0}+1)^{1/2}a_{0}