What is the tangent of y=sin(2x)ln(x^2),(1,0) ?

The tangent of y=sin(2x)ln(x^2),(1,0) is y=2sin(2)x-2sin(2)

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tangent of y=sin(2x)ln(x^2),(1,0)

tangent of

y = sin (2 x) ln (x^{2}), (1, 0)

y = 2 sin (2) x - 2 sin (2)

Solution steps

Find the slope of

y = sin (2 x) ln (x^{2}) : \frac{d y}{d x} = 2 cos (2 x) ln (x^{2}) + \frac{2 sin ( 2 x )}{x}

Compute the slope of

y = sin (2 x) ln (x^{2})

(1, 0) : m = 2 sin (2)

Find the line with slope m=

2 sin (2)

and passing through

(1, 0) : y = 2 sin (2) x - 2 sin (2)

y = 2 sin (2) x - 2 sin (2)

\int \frac{v}{1 + v ^{2}} d v

\int sinh (x) cosh (x) d x

1 - x, 0

y^{^{'''}} + 3 y^{^{''}} - 4 y = 0

What is the tangent of y=sin(2x)ln(x^2),(1,0) ?
The tangent of y=sin(2x)ln(x^2),(1,0) is y=2sin(2)x-2sin(2)