What is (d^2)/(dx^2)(sin(bx)-2cos(bx)) ?

The answer to (d^2)/(dx^2)(sin(bx)-2cos(bx)) is -b^2sin(bx)+2b^2cos(bx)

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(d^2)/(dx^2)(sin(bx)-2cos(bx))

\frac{d ^{2}}{d x ^{2}} (sin (b x) - 2 cos (b x))

- b^{2} sin (b x) + 2 b^{2} cos (b x)

Solution steps

\frac{d ^{2}}{d x ^{2}} (sin (b x) - 2 cos (b x))

\frac{d}{d x} (sin (b x) - 2 cos (b x)) = b cos (b x) + 2 b sin (b x)

= \frac{d}{d x} (b cos (b x) + 2 b sin (b x))

\frac{d}{d x} (b cos (b x) + 2 b sin (b x)) = - b^{2} sin (b x) + 2 b^{2} cos (b x)

= - b^{2} sin (b x) + 2 b^{2} cos (b x)

\int \frac{1}{b x} d x

h \to 0 lim ((1 + h)^{\frac{1}{h}})

n = 1 \sum \infty n^{- \frac{1}{2}}

y = \frac{x ^{4}}{1 - x ^{3}}

x \to 5 + lim (\frac{x - 6}{x - 5})

What is (d^2)/(dx^2)(sin(bx)-2cos(bx)) ?
The answer to (d^2)/(dx^2)(sin(bx)-2cos(bx)) is -b^2sin(bx)+2b^2cos(bx)