What is the derivative of y=(2x+1)^{sin(5x)} ?

The derivative of y=(2x+1)^{sin(5x)} is (5cos(5x)ln(2x+1)+(2sin(5x))/(2x+1))(2x+1)^{sin(5x)}

What is the first derivative of y=(2x+1)^{sin(5x)} ?

The first derivative of y=(2x+1)^{sin(5x)} is (5cos(5x)ln(2x+1)+(2sin(5x))/(2x+1))(2x+1)^{sin(5x)}

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derivative of

y = (2 x + 1)^{s i n (5 x)}

(5 cos (5 x) ln (2 x + 1) + \frac{2 sin ( 5 x )}{2 x + 1}) (2 x + 1)^{s i n (5 x)}

Solution steps

\frac{d}{d x} ((2 x + 1)^{s i n (5 x)})

Apply exponent rule:

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

(2 x + 1)^{s i n (5 x)} = e^{s i n (5 x) l n (2 x + 1)}

= \frac{d}{d x} (e^{s i n (5 x) l n (2 x + 1)})

Apply the chain rule:

e^{s i n (5 x) l n (2 x + 1)} \frac{d}{d x} (sin (5 x) ln (2 x + 1))

= e^{s i n (5 x) l n (2 x + 1)} \frac{d}{d x} (sin (5 x) ln (2 x + 1))

\frac{d}{d x} (sin (5 x) ln (2 x + 1)) = 5 cos (5 x) ln (2 x + 1) + \frac{2 sin ( 5 x )}{2 x + 1}

= e^{s i n (5 x) l n (2 x + 1)} (5 cos (5 x) ln (2 x + 1) + \frac{2 sin ( 5 x )}{2 x + 1})

e^{s i n (5 x) l n (2 x + 1)} = (2 x + 1)^{s i n (5 x)}

= (5 cos (5 x) ln (2 x + 1) + \frac{2 sin ( 5 x )}{2 x + 1}) (2 x + 1)^{s i n (5 x)}

\frac{5}{3 x ^{2}}

y = lo g_{10} (3 x)

f (x) = (1 + 2 e^{x}) sin (x)

- 2^{x}

\frac{1}{4} \cdot x^{2} - \frac{( sin ( 2 x ^{2} ) )}{8}

What is the derivative of y=(2x+1)^{sin(5x)} ?
The derivative of y=(2x+1)^{sin(5x)} is (5cos(5x)ln(2x+1)+(2sin(5x))/(2x+1))(2x+1)^{sin(5x)}
What is the first derivative of y=(2x+1)^{sin(5x)} ?
The first derivative of y=(2x+1)^{sin(5x)} is (5cos(5x)ln(2x+1)+(2sin(5x))/(2x+1))(2x+1)^{sin(5x)}