What is derivative of x=(t^2+1)^{1/2} ?

The answer to derivative of x=(t^2+1)^{1/2} is (dt}{dx}=\frac{(t^2+1)^{1/2})/t

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derivative of x=(t^2+1)^{1/2}

derivative of

x = (t^{2} + 1)^{\frac{1}{2}}

\frac{d t}{d x} = \frac{( t ^{2} + 1 ) ^{\frac{1}{2}}}{t}

Solution steps

x = (t^{2} + 1)^{\frac{1}{2}}

Treat

t

t (x)

Differentiate both sides:

1 = \frac{t \frac{d t}{d x}}{( t ^{2} + 1 ) ^{\frac{1}{2}}}

Isolate

\frac{d t}{d x} : \frac{d t}{d x} = \frac{( t ^{2} + 1 ) ^{\frac{1}{2}}}{t}

\frac{d t}{d x} = \frac{( t ^{2} + 1 ) ^{\frac{1}{2}}}{t}

x + y - x^{3} y = 15

x = (1 - y^{2})^{\frac{1}{2}}

2 x^{2} + 2 y - 2 x y = 4

x^{2} + 2 x y = y^{2}

x^{3} + x^{2} y + 7 x y^{2} + y^{3} = 81

What is derivative of x=(t^2+1)^{1/2} ?
The answer to derivative of x=(t^2+1)^{1/2} is (dt}{dx}=\frac{(t^2+1)^{1/2})/t