What is derivative of x=e^{sin(t^2-4t)} ?

The answer to derivative of x=e^{sin(t^2-4t)} is (dt)/(dx)= 1/(2e^{sin(t^2-4t))cos(t^2-4t)(t-2)}

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derivative of x=e^{sin(t^2-4t)}

derivative of

x = e^{s i n (t^{2} - 4 t)}

\frac{d t}{d x} = \frac{1}{2 e ^{s i n (t^{2} - 4 t)} cos ( t ^{2} - 4 t ) ( t - 2 )}

Solution steps

x = e^{s i n (t^{2} - 4 t)}

Treat

t

t (x)

Differentiate both sides:

1 = e^{s i n (t^{2} - 4 t)} cos (t^{2} - 4 t) (2 t \frac{d t}{d x} - 4 \frac{d t}{d x})

Isolate

\frac{d t}{d x} : \frac{d t}{d x} = \frac{1}{2 e ^{s i n (t^{2} - 4 t)} cos ( t ^{2} - 4 t ) ( t - 2 )}

\frac{d t}{d x} = \frac{1}{2 e ^{s i n (t^{2} - 4 t)} cos ( t ^{2} - 4 t ) ( t - 2 )}

(d^{2} + 1) \cdot y = x^{2} \cdot (cos (x))

x^{3} + y^{3} = 9

x^{2} + y^{2} - 3 x y = 1

7 x^{4} - 4 y = 4

4 x + 5 y = lo g_{10} (4 x - 3 y)

What is derivative of x=e^{sin(t^2-4t)} ?
The answer to derivative of x=e^{sin(t^2-4t)} is (dt)/(dx)= 1/(2e^{sin(t^2-4t))cos(t^2-4t)(t-2)}