Question

# Differentiate with respect to$\theta$:$\theta {x^2} + 9{y^2}$

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
$\left( {\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) + g\left( x \right) = \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) + \dfrac{d}{{dx}}\left( {g\left( x \right)} \right)} \right)$