What is the derivative of sqrt(e^{1+x^2)+3x^4} ?

The derivative of sqrt(e^{1+x^2)+3x^4} is (x(6x^2+e^{x^2+1}))/(sqrt(e^{x^2+1)+3x^4)}

What is the first derivative of sqrt(e^{1+x^2)+3x^4} ?

The first derivative of sqrt(e^{1+x^2)+3x^4} is (x(6x^2+e^{x^2+1}))/(sqrt(e^{x^2+1)+3x^4)}

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derivative of

e^{1 + x^{2}} + 3 x^{4}

\frac{x ( 6 x ^{2} + e ^{x^{2} + 1} )}{e ^{x^{2} + 1} + 3 x ^{4}}

Solution steps

\frac{d}{d x} (e^{1 + x^{2}} + 3 x^{4})

Apply the chain rule:

\frac{1}{2 e ^{1 + x^{2}} + 3 x ^{4}} \frac{d}{d x} (e^{1 + x^{2}} + 3 x^{4})

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

\frac{d}{d x} (e^{1 + x^{2}} + 3 x^{4}) = e^{1 + x^{2}} \cdot 2 x + 12 x^{3}

= \frac{1}{2 e ^{1 + x^{2}} + 3 x ^{4}} (e^{1 + x^{2}} \cdot 2 x + 12 x^{3})

Simplify

\frac{1}{2 e ^{1 + x^{2}} + 3 x ^{4}} (e^{1 + x^{2}} \cdot 2 x + 12 x^{3}) : \frac{x ( 6 x ^{2} + e ^{x^{2} + 1} )}{e ^{x^{2} + 1} + 3 x ^{4}}

= \frac{x ( 6 x ^{2} + e ^{x^{2} + 1} )}{e ^{x^{2} + 1} + 3 x ^{4}}

\frac{d y}{d x} y = sin (- 1 (6 x + 1))

\int 10 x 3 x^{2} - 4 d x

\frac{\partial}{\partial y} (x^{2} + 2 y^{2} - x^{2} y)

x \to \frac{7 π}{2} + lim (sec (x))

\frac{d}{d x} ((2 x - 9)^{4} (x^{2} + x + 1)^{5})

What is the derivative of sqrt(e^{1+x^2)+3x^4} ?
The derivative of sqrt(e^{1+x^2)+3x^4} is (x(6x^2+e^{x^2+1}))/(sqrt(e^{x^2+1)+3x^4)}
What is the first derivative of sqrt(e^{1+x^2)+3x^4} ?
The first derivative of sqrt(e^{1+x^2)+3x^4} is (x(6x^2+e^{x^2+1}))/(sqrt(e^{x^2+1)+3x^4)}