If y = 5\sin x + e^{3x}, what is the value of \frac{dy}{dx} at x = 0? — JAMB UTME

If $y = 5\sin x + e^{3x}$ , what is the value of $\frac{dy}{dx}$ at $x = 0$ ?

Answer: 8

The value of the derivative at

x = 0

is 8.

Differentiating

y = 5\sin x + e^{3x}

yields

\frac{dy}{dx} = 5\cos x + 3e^{3x}

. Substituting

x = 0

gives

5\cos(0) + 3e^{0} = 5(1) + 3(1) = 8

Differentiate each term of the function

y = 5\sin x + e^{3x}

with respect to

x

\frac{dy}{dx} = 5\cos x + 3e^{3x}

The derivative of

\sin x

\cos x

, and applying the chain rule to

e^{3x}

gives

3e^{3x}

Evaluate the derivative

\frac{dy}{dx}

x = 0

5\cos(0) + 3e^0 = 5(1) + 3(1) = 8

Evaluating trigonometric and exponential functions at zero gives

\cos(0) = 1

and

e^0 = 1

Differentiation of trigonometric and exponential functions and evaluation of derivatives at specific points.

Question