What is the derivative of 4^{sin^2(x+cos(3x)}) ?

The derivative of 4^{sin^2(x+cos(3x)}) is ln(2)*2^{2(sin^2(x)+cos(3x))+1}(sin(2x)-3sin(3x))

What is the first derivative of 4^{sin^2(x+cos(3x)}) ?

The first derivative of 4^{sin^2(x+cos(3x)}) is ln(2)*2^{2(sin^2(x)+cos(3x))+1}(sin(2x)-3sin(3x))

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\frac{d}{d x} (4^{s i n^{2} (x) + c o s (3 x)})

ln (2) \cdot 2^{2 (s i n^{2} (x) + c o s (3 x)) + 1} (sin (2 x) - 3 sin (3 x))

Solution steps

\frac{d}{d x} (4^{s i n^{2} (x) + c o s (3 x)})

Apply exponent rule:

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

4^{s i n^{2} (x) + c o s (3 x)} = e^{(s i n^{2} (x) + c o s (3 x)) l n (4)}

= \frac{d}{d x} (e^{(s i n^{2} (x) + c o s (3 x)) l n (4)})

Apply the chain rule:

e^{(s i n^{2} (x) + c o s (3 x)) l n (4)} \frac{d}{d x} ((sin^{2} (x) + cos (3 x)) ln (4))

= e^{(s i n^{2} (x) + c o s (3 x)) l n (4)} \frac{d}{d x} ((sin^{2} (x) + cos (3 x)) ln (4))

\frac{d}{d x} ((sin^{2} (x) + cos (3 x)) ln (4)) = 2 ln (2) (sin (2 x) - sin (3 x) \cdot 3)

= e^{(s i n^{2} (x) + c o s (3 x)) l n (4)} \cdot 2 ln (2) (sin (2 x) - sin (3 x) \cdot 3)

Simplify

e^{(s i n^{2} (x) + c o s (3 x)) l n (4)} \cdot 2 ln (2) (sin (2 x) - sin (3 x) \cdot 3) : ln (2) \cdot 2^{2 (s i n^{2} (x) + c o s (3 x)) + 1} (sin (2 x) - 3 sin (3 x))

= ln (2) \cdot 2^{2 (s i n^{2} (x) + c o s (3 x)) + 1} (sin (2 x) - 3 sin (3 x))

2 ln (x) + 6

\frac{\partial}{\partial x} (x y - x - y + 1)

\int \frac{1}{y ^{2} - y + 1} d y

\int_{0}^{1} 7^{2 x} d x

What is the derivative of 4^{sin^2(x+cos(3x)}) ?
The derivative of 4^{sin^2(x+cos(3x)}) is ln(2)*2^{2(sin^2(x)+cos(3x))+1}(sin(2x)-3sin(3x))
What is the first derivative of 4^{sin^2(x+cos(3x)}) ?
The first derivative of 4^{sin^2(x+cos(3x)}) is ln(2)*2^{2(sin^2(x)+cos(3x))+1}(sin(2x)-3sin(3x))