What is the derivative of 3^{x^4-2x^2} ?

The derivative of 3^{x^4-2x^2} is ln(3)*3^{x^4-2x^2}(4x^3-4x)

What is the first derivative of 3^{x^4-2x^2} ?

The first derivative of 3^{x^4-2x^2} is ln(3)*3^{x^4-2x^2}(4x^3-4x)

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\frac{d}{d x} (3^{x^{4} - 2 x^{2}})

ln (3) \cdot 3^{x^{4} - 2 x^{2}} (4 x^{3} - 4 x)

Solution steps

\frac{d}{d x} (3^{x^{4} - 2 x^{2}})

Apply exponent rule:

a^{b} = e^{b l n (a)}

3^{x^{4} - 2 x^{2}} = e^{(x^{4} - 2 x^{2}) l n (3)}

= \frac{d}{d x} (e^{(x^{4} - 2 x^{2}) l n (3)})

Apply the chain rule:

e^{(x^{4} - 2 x^{2}) l n (3)} \frac{d}{d x} ((x^{4} - 2 x^{2}) ln (3))

= e^{(x^{4} - 2 x^{2}) l n (3)} \frac{d}{d x} ((x^{4} - 2 x^{2}) ln (3))

\frac{d}{d x} ((x^{4} - 2 x^{2}) ln (3)) = ln (3) (4 x^{3} - 4 x)

= e^{(x^{4} - 2 x^{2}) l n (3)} ln (3) (4 x^{3} - 4 x)

e^{(x^{4} - 2 x^{2}) l n (3)} = 3^{x^{4} - 2 x^{2}}

= ln (3) \cdot 3^{x^{4} - 2 x^{2}} (4 x^{3} - 4 x)

x \to 0 lim ((5 x - 3)^{2})

\frac{( 2 s + 1 )}{( s + 1 ) ^{2}}

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

e^{w}

x^{3} - 2 x^{2} - 5 x + 6, at x = 2

What is the derivative of 3^{x^4-2x^2} ?
The derivative of 3^{x^4-2x^2} is ln(3)*3^{x^4-2x^2}(4x^3-4x)
What is the first derivative of 3^{x^4-2x^2} ?
The first derivative of 3^{x^4-2x^2} is ln(3)*3^{x^4-2x^2}(4x^3-4x)