Question

Differentiate with respect to\[\theta \]:
\[\theta {x^2} + 9{y^2}\]

Answer 1

Hint: It is easily solved by simple derivation with respect to \[\theta \]. Except x and y variables are constant so, x we will differentiate $\theta $ only.


Complete step by step solution:
Let \[y' = \theta {x^2} + 9{y^2}\]
Now, differentiate with respect\[\theta \].So y and x are constant terms.
\[\dfrac{{dy'}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {\theta {x^2} + 9{y^2}} \right)\]
\[ \Rightarrow \dfrac{{d{y^,}}}{{d\theta }} = \,\dfrac{d}{{d\theta }}\left( {\theta {x^2}} \right) + \dfrac{d}{{d\theta }}\left( {9{y^2}} \right)\] \[\left[ {\because \,\,\dfrac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right)\,\, = \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) + \dfrac{d}{{dx}}\left( {g\left( x \right)} \right)} \right]\] \[ \Rightarrow \dfrac{{d{y^,}}}{{d\theta }} = \,{x^2}\dfrac{d}{{d\theta }}\left( \theta \right) + 9{y^2}\dfrac{d}{{d\theta }}\left( 1 \right)\]
 \[ \Rightarrow \dfrac{{dy}}{{d\theta }} = \,1 \times {x^2} + 9{y^2}.0\]
\[ \Rightarrow \dfrac{{dy}}{{d\theta }} = \,1 \times {x^2} + 0\]
\[
   \Rightarrow \dfrac{{d{y^,}}}{{d\theta }}\, = {x^2} + 0 \\
   \Rightarrow \dfrac{{d{y^,}}}{{d\theta }} = {x^2} \\
 \]
Additional information: Differentiation is a process of finding the derivative, or rate of change, of a function. Differentiation rules:
(i) The constant rule: for any fixed real number $c$.\[\dfrac{d}{{dx}}\left\{ {c.f(x)} \right\} = c.\dfrac{d}{{dx}}\left\{ {f(x)} \right\}\]
(ii) The power rule: $\dfrac{d}{{dx}}\left\{ {{x^n}} \right\} = n{x^{n - 1}}$


Note: We can solve similar question by simple derivation as
\[\left( {\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) + g\left( x \right) = \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) + \dfrac{d}{{dx}}\left( {g\left( x \right)} \right)} \right)\]